\(\frac { 1 }{ 2 } \)

Given equation of line is 3x - 2y + 6 = 0
2y = 3x + 6 \(\Rightarrow\) y = \(\frac{3x+6}{2}\)

x	0	-2
t	3	0

\(\therefore Area=\int _{ -3 }^{ -2 }{ -ydx+ydx } +\int _{ -2 }^{ 1 }{ ydx } \)
[\(\because\) the Area is below the x - axis]

\(=\frac { -1 }{ 2 } \int _{ -3 }^{ -2 }{ (3x+6)dx+\frac { 1 }{ 2 } \int _{ -2 }^{ 1 }{ (3x+6)dx } } \)
\(={ \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -3 }^{ -2 }+\frac { 1 }{ 2 } { \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -2 }^{ 1 }\)
\(=-\frac { 1 }{ 2 } \left[ \left( \frac { 12 }{ 2 } -12 \right) -\left( \frac { 27 }{ 2 } -18 \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3 }{ 2 } +6 \right) -\left( \frac { 12 }{ 2 } -12 \right) \right] \)
\(=-\frac { 1 }{ 2 } \left[ (-6)-\left( \frac { 27-36 }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3+12 }{ 2 } \right) -(-6) \right] \)
\(=-\frac { 1 }{ 2 } \left[ -6+\frac { 9 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 15 }{ 2 } +6 \right] \)
\(\\ =-\frac { 1 }{ 2 } \left[ \frac { -3 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 27 }{ 2 } \right] =\frac { 3 }{ 4 } +\frac { 27 }{ 4 } =\frac { 30 }{ 4 } =\frac { 15 }{ 2 } \)
\(\therefore\)  A = 7.5 sq.units

\( \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -ax }dx}=\frac { n! }{ { a }^{ n+1 } } \)
\(n=5,\quad a=3 \)
\(=\frac { 5! }{ { 3 }^{ 6 } } \)

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ n }x } =\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2\)
\(Let\quad { I }_{ 10 }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 10 }xdx=\frac { 9 }{ 10 } { I }_{ 8 } } \)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times { I }_{ 6 }=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times { I }_{ 4 }\)
\(\\ =\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } { I }_{ 2 }\)
\(=\frac { 9 }{ 10 } \times \frac { 7 }{ 8 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } \)
\(=\frac { 315 }{ 1280 } \times \frac { \pi }{ 2 } =\frac { 63\pi }{ 256(2) } =\frac { 63\pi }{ 512 } \)

Given that I =\(\int ^\frac {\pi}{2}_{0} \)( sin²x + cos⁴x)dx =\(\int ^\frac {\pi}{2}_{0} \) sin²x dx+\(\int ^\frac {\pi}{2}_{0} \)cos⁴x dx\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)

\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } \)
\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } =\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| +c\right] \)
\(=\frac { 1 }{ 4 } \left[ log\left( \frac { 4-2 }{ 4+2 } \right) -log\left( \frac { 3-2 }{ 3+2 } \right) \right] \)
\(=\frac { 1 }{ 4 } log\left[ \left( \frac { 2 }{ 6 } \right) - log \ \frac { 1 }{ 5 } \right] \\ =\frac { 1 }{ 4 } log\left( \frac { 1 }{ 3 } \times 5 \right) \)
\(=\frac { 1 }{ 4 } log\left( \frac { 5 }{ 3 } \right) \)

Let f (x) = x cos x
Then f (−x) = (−x) cos(−x) = −x cos x = − f (x).
So f (x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x), \(\int _{ -a }^{ a }{ f(x)dx=0 } \)
\(\therefore\) we get \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

Take 2a = 2\(\pi\) and f(x) = g(cosx)
Then, f (2a−x) = f(2\(\pi\)-x) = g(cos(2\(\pi\)-x)) = g(cos x) = f(x)
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=2 } \int _{ 0 }^{ a }{ f(x)dx } \)
\(\therefore \int _{ 0 }^{ 2\pi }{ g(cosx)dx=2\int _{ 0 }^{ \pi }{ g(cosx)dx } } \)

\(\int _{ 0 }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ [2x] } dx+\int _{ \frac { 1 }{ 2 } }^{ 1 }{ [2x]dx } =\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 0dx+ } \int _{ \frac { 1 }{ 2 } }^{ 1 }{ 1 dx} = 0+[x]^1_{\frac{1}{2}} = 1 -\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)

The region to be revolved about the x-axis is sketched as in

\(V=\pi \int_{0}^{1}y^{2}dx=\pi \int_{0}^{1}\left ( x^{2}+4x+5 \right )^{2}dx\)
\(=\pi \int_{0}^{1}\left ( x^{4}+16x^{2} +25+8x^{3}+40x+10x^{2}\right )dx\)
\(=\pi \left ( \frac{x^{5}}{5}+8\frac{x^{4}}{4}+26\frac{x^{3}}{3}+40\frac{x^{2}}{2}+25x \right )_{0}^{1}\)
\(=\pi \left ( \frac{1}{5}+2+\frac{26}{3}+20+25 \right )=\frac{838}{15}\pi \)

c=-1

Given x = t², y = t - \(\frac { { t }^{ 3 } }{ 3 } \) 
Point of intersection of the curve with x-axis is obtained by putting y = 0.
∴ y = 0
⇒ \(t-\frac { { t }^{ 3 } }{ 3 } =0\)
⇒ t = 0 or ± \(\sqrt3\)
∴ The limit is from t = 0 to t = \(\sqrt { 3 } \)
∴ Volume = \(\pi \int _{ 0 }^{ \sqrt { 3 } }{ { y }^{ 2 }dx } =\pi \int _{ 0 }^{ \sqrt { 3 } }{ { \left( t-\frac { { t }^{ 3 } }{ 3 } \right) }^{ 2 }}(2t \ dt)\)
[∵ x = t² ⇒ dx = 2t dt]
\(=2\pi \int _{ 0 }^{ \sqrt { 3 } }{ { \left( \frac { 3t-{ t }^{ 3 } }{ 3 } \right) }^{ 2 }t \ dt } \)
\(=\frac { 2\pi }{ 9 } \int _{ 0 }^{ \sqrt { 3 } }{ (9{ t }^{ 2 }-6{ t }^{ 4 }+{ t }^{ 6 })t \ dt } \)
\(=\frac { 2\pi }{ 9 } \int _{ 0 }^{ \sqrt { 3 } }{ (9{ t }^{ 3 }-6{ t }^{ 5 }+{ t }^{ 7 })dt } \)
\(=\frac { 2\pi }{ 9 } { \left[ \frac { { 9 }t^{ 4 } }{ 4 } -\frac { 6{ t }^{ 6 } }{ 6 } +\frac { { t }^{ 8 } }{ 8 } \right] }_{ 0 }^{ \sqrt { 3 } }\)
\(=\frac { 2\pi }{ 9 } \left[ \frac { 81 }{ 4 } -27+\frac { 81 }{ 8 } -0 \right] \)
Volume \(=\frac { 2\pi }{ 9 } \left( \frac { 27 }{ 8 } \right) =\frac { 3\pi }{ 4 } \) cubic units

Volume of the right circular conical frustum is obtained by revolving the line y = x between x = a and x = b around the x - axis

\(\therefore\) Height of the frustum h = b - a
\(\therefore\)Volume \(=\pi \int _{ a }^{ b }{ { x }^{ 2 }dx } =\pi { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ a }^{ b }\)
\(=\frac { \pi }{ 3 } [{ b }^{ 3 }-{ a }^{ 3 }]\)
\(=\frac { \pi }{ 3 } (b-a)({ b }^{ 2 }+ab+{ a }^{ 2 })\)
Now, substitute h = b - a, r = a and R = b we get Volume of the conical frustum
\(\frac { \pi }{ 3 } [h({ R }^{ 2 }+rR+{ r }^{ 2 })]\)
Given h = 2 m, r = 1 m, R = 2 m we get
Required volume \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

Equation of the given curve is y + 1 = x²
\(\therefore\) The vertex of this open upward parabola is (0, -1)

Required volume \(=\int _{ -1 }^{ 3 }{ { x }^{ 2 }dy } \)
\(=\pi \int _{ -1 }^{ 3 }{ (1+y)dy } \)
\(=\pi { \left[ y+\frac { { y }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 3 }=\pi \left[ \left( 3+\frac { 9 }{ 2 } \right) -\left( -1+\frac { 1 }{ 2 } \right) \right] \)
\(=\pi \left[ \left( \frac { 6+9 }{ 2 } \right) -\left( -\frac { 1 }{ 2 } \right) =\pi \left( \frac { 15 }{ 2 } +\frac { 1 }{ 2 } \right) =\left( \frac { 16 }{ 2 } \right) \right] \)
V = 8\(\pi\)

Equation of the given curve is y = e^-2x
Required Volume = \(\pi \int _{ 0 }^{ 1 }{ { { (e }^{ -2x }) }^{ 2 }dx } \)
\(=\pi \int _{ 0 }^{ 1 }{ { e }^{ -4x } } dx=\pi { \left[ \frac { { e }^{ -4x } }{ -4 } \right] }_{ 0 }^{ 1 }\)
\(=\frac { -\pi }{ 4 } \left[ { e }^{ -4 }-{ e }^{ -0 } \right] =-\frac { \pi }{ 4 } \left( { e }^{ -4 }-1 \right) \)
\(V=\frac { \pi }{ 4 } (1-{ e }^{ -4 })\) cubic units

Consider the triangular region in the first quadrant which is bounded by the line y = \(\frac{r}{h}\)x. x-axis, the lines x = 0 and x = h. revolving the region about the x-axis, we get a cone of base radius r and height h.

Hence, the volume of the cone is given by
\(\\ \\ V=\pi \int _{ 0 }^{ h }{ { y }^{ 2 }dx=\pi \int _{ 0 }^{ h }{ { \left( \frac { r }{ h } x \right) }^{ 2 } } dx=\pi { \left( \frac { r }{ h } \right) }^{ 2 } } { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ h }=\frac { \pi { r }^{ 2 }h }{ 3 } \)

Given equation of the parabola is (y-1) = (x-2)²
\(\Rightarrow\) y = (x - 2)² + 1
It vertex is (2, 1) which is the minimum point P. Let Q(x, y) be a point on the parabola given slope of PQ = 2
\(\Rightarrow \frac { y-1 }{ x-2 } =2\ \left[ \because slope=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\(\Rightarrow\) y-1 = 2(x-2) \(\Rightarrow\) y-1 = 2x-4 \(\Rightarrow\) y = 2x-4+1
\(\Rightarrow\) y = 2x + 3
From (1) and (2), (x-2)²+1 = 2x-3
\(\Rightarrow\) x²-4x + 4 + 1 = 2x - 3 \(\Rightarrow\) x²- 6x + 8 = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4

\(\therefore\) Required area \(=\int _{ 2 }^{ 4 }{ ({ y }_{ 1 }-{ y }_{ 2 }) } dx\)
\(=\int _{ 2 }^{ 4 }{ (2x-3)-{ (x-2) }^{ 2 }-1dx } \)
\(=\int _{ 2 }^{ 4 }{ (2x-3-{ x }^{ 2 }+4x-4-1)dx } \)
\(=\int _{ 2 }^{ 4 }{ (-{ x }^{ 2 }+6x-8)dx } \)
\({ \left[ \frac { -{ x }^{ 3 } }{ 3 } +3{ x }^{ 2 }-8x \right] }_{ 2 }^{ 4 }=\left( \frac { -64 }{ 3 } +48-32 \right) -\left( \frac { -8 }{ 3 } +12-16 \right) \)
\(=\left( \frac { -64 }{ 3 } +16 \right) -\left( -\frac { 8 }{ 3 } -4 \right) \)
\(=\left( \frac { -64+48 }{ 3 } \right) -\left( \frac { -8-12 }{ 3 } \right) \)
\(=\frac { 16 }{ 3 } \) sq.units

The required area is sketched. One portion of the region lies above the x−axis between x = 0 and x = \(\pi\), and the other portion lies below x−axis between x = \(\pi\) and x = 2\(\pi\). So, the required area is given by 

The ellipse is symmetric about both major and minor axes. It is sketched. So, viewing in the positive direction of y -axis, the required area A is four times the area of the region bounded by the portion of the ellipse in the first quadrant 
\(\left( y=\frac { b }{ a } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } ,0

x -axis, x = 0 and x = a .
Hence, by taking vertical strips, we get \(A=4\int _{ 0 }^{ a }{ ydx=4\int _{ 0 }^{ a }{ \frac { b }{ a } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \)
\(=\frac { 4b }{ a } { \left[ \frac { x\sqrt { { a }^{ 2 }-{ y }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ 0 }^{ a }=\frac { 4b }{ a } \times \frac { \pi { a }^{ 2 } }{ 4 } =\pi ab\)

\(=\int _{ 0 }^{ \pi /2 }{ { e }^{ -tan\quad x }{ sec }^{ 6 }xdx\ \left[ \because sec\ x=\frac { 1 }{ cos\quad x } \right] } \)
\(Put\quad t=tan\quad x\Rightarrow dt={ sec }^{ 2 }xdx\)
\(=\int _{ 0 }^{ \pi /2 }{ { e }^{ -tanx }{ sec }^{ 4x }{ sec }^{ 2x }dx } \)
\(=\int _{ 0 }^{ \pi /2 }{ { e }^{ -tanx }{ (1+{ tan }^{ 2 }) }^{ 2 } } { sec }^{ 2 }xdx\)
\(put\quad t=tanx\Rightarrow dt={ sec }^{ 2 }xdx\)

x	0	\(\frac{\pi}{2}\)
t	0	\(\infty\)

\(=\int _{ 0 }^{ \infty }{ { e }^{ -t }{ (1+{ t }^{ 2 }) }^{ 2 } } dt\)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -t }(1+{ t }^{ 4 }+{ 2t }^{ 2 })dt } \)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -t }(1)dt+\int _{ 0 }^{ \infty }{ { e }^{ -t }{ t }^{ 4 }dt+2 } } \int _{ 0 }^{ \infty }{ { e }^{ -t }{ t }^{ 2 }dt } \)
\(=0!+4!+2(2!)\left[ \because \int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx=n! } \right] \)
\(=1+4\times 3\times 2\times 1+2\times 2=1+24+4\)
\(\therefore I=29\)

Let I =\(\int ^{\pi}_{0} \frac{x}{1+sin x}\)
= \(\int ^{\pi}_{0} x \frac{x}{1+sin x}\)dx
Let f (x) = \(\frac{1}{1+ sin x}\) 
Then f(π-x) = \(\frac{1}{1+ sin (\pi - x)}\) = \(\frac{1}{1+ sin x}\) = f(x)
∴  \(\int ^{\pi}_{0} \frac{x}{1+sin x}\) dx = \(\frac{\pi}{2}\)  \(\int ^{\pi}_{0} \frac{1}{1+sin x}\) dx,      \((\because \int _{ 0 }^{ a }{ xf(x)dx=\frac { a }{ 2 } \int _{ 0 }^{ a }{ f(x)dx } if } f(a-x)=f(x))\)
= \(\pi\) \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+sin x}\) dx,       (∴  \(\int ^{\pi}_0\) g(sin x)dx = 2  \(\int ^{ \frac{\pi}{2}}_0\) g(sin x)dx)
= \(\pi\)  \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+sin (\frac{\pi}{2} -x)}\) dx  (∴ \(\int ^{a}_{0} \) f(x) dx =  \(\int ^{a}_{0} \) f(a-x)dx)
= \(\pi\)  \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+cos x}\) dx = \(\pi\) \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{2cos^2 \frac{x}{2}}\) dx = \(\frac{\pi}{2}\) \(\int ^{ \frac{\pi}{2}}_0 sec ^2 \frac{x}{2}\) dx
= \(\pi\)  \([tan \frac{x}{2}]^\frac{\pi}{2}_0\) =  \(\pi\) \([tan \frac{\pi}{4} - tan 0]\) = \(\pi\)

\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ sin^{ 4 }x+cos^{ 4 }x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (sin^{ 2 }x+cos^{ 2 }x)-2sin^{ 2 }xcos^{ 2 }x } \)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (1-\frac { 1 }{ 2 } (2sinxcosx)^{ 2 } } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 2-sin^{ 2 }2x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 1+cos^{ 2 }2x } \)
Put u = cos 2x, Then, du = −2sin 2x dx
When x = 0 , we have u = cos 0 = 1. When x = \(\frac{\pi}{4}\) we have u = cos\(\frac{\pi}{4}\) = 0
∴ I =  \(\int^{0}_{1} \frac {-du}{1+u^2}\) = \(\int^{0}_{1} \frac {du}{1+u^2}\) = [tan^-1 u\(]^1_0\) = \(\frac{\pi}{4}\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos\theta }{ (1+sin\theta )(2+sin\theta ) } } d\theta \)
Put u = 1 + sin\(\theta\)
Then, du = cos\(\theta\) d\(\theta\)
When \(\theta\) = 0, u = 1 
When \(\theta =\frac{\pi}{2}, u=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ u(1+u) } } =\int _{ 1 }^{ 2 }{ \frac { (1+u)-u }{ u(1+u) } du } =\int _{ 1 }^{ 2 }{ \left( \frac { 1 }{ u } -\frac { 1 }{ 1+u } \right) du=[logu-log(1+u)]_{ 1 }^{ 2 } } \)
\(=(log2-log3)-(log1-log2)=2log2-log3=log\frac { 4 }{ 3 } .\)

Equation of the given curve is y = 2x²
Volume \(=\pi \int _{ a }^{ b }{ { y }^{ 2 }dx } \)
\(=\pi \int _{ 0 }^{ 1 }{ { (2{ x }^{ 2 }) }^{ 2 } } dx=\pi \int _{ 0 }^{ 1 }{ { 4x }^{ 4 }dx } \)
\(=4\pi { \left[ \frac { { x }^{ 5 } }{ 5 } \right] }_{ 0 }^{ 1 }=\frac { 4\pi }{ 5 } (1-0)\)
\(v=\frac { 4\pi }{ 5 } \) cubic units

Making the substitution t = ax, we get dt = adx and x = 0 \(\Rightarrow\) t = 0 and x = \(\infty\) \(\Rightarrow\) t = \(\infty\)
Hence, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -ax }{ x }^{ n }dx } =\int _{ 0 }^{ \infty }{ { e }^{ -t } } { \left( \frac { t }{ a } \right) }^{ n }\frac{dt}{a}= \int^\infty_0 e^{-t }t^n dt\)
\(=\frac { 1 }{ { a }^{ n+1 } } \int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx } =\frac { n! }{ { a }^{ n+1 } } \)
Thus
\(\int _{ 0 }^{ \infty }{ { e }^{ -ax }{ x }^{ n }dx } =\frac { n! }{ { a }^{ n+1 } } \)

\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 3 }\theta { cos }^{ 5 }\theta d\theta } \)
\( \mathrm{m}=3, \mathrm{n}=5 \)
\( \int_{0}^{\pi / 2} (cos ^5 \theta - cos^7 \theta) sin \theta d\theta\)
 \(t = cos \theta\)
\(dt = -sin \theta d\theta\)
\(= \int ^0_1(t^2-t^7)(-dt)\\ = \int ^0_1(t^5-t^7)(dt)= [\frac{t^6}{6}-\frac{t^8}{8}]\)
\(\frac{1}{6}-\frac{1}{8}= \frac{8-6}{48}\)
\( =\frac{1}{24} \)
Aliter method:
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 3 }\theta { cos }^{ 5 }\theta d\theta } \)
Here m = 3, which is odd and n = 5, which is odd
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ m }\theta { cos }^{n } }x dx \)
\( \frac{n-1}{m+n} \cdot \frac{n-3}{m+n-2} \cdot \frac{n-5}{m+n-4} \cdots \frac{2}{m+3} \cdot \frac{1}{m+1} \\ \)
\( \int_{0}^{\pi / 2} \sin ^{3} \theta \cos ^{5} \theta d \theta=\frac{A}{8} \times \frac{2}{6} \times \frac{1}{A} \\ =\frac{1}{24} \)

\(Let\ I=\int _{ 0 }^{ \pi /2 }{ { sin }^{ 2 }x{ cos }^{ 4 }xdx } \)
\({ I }_{ m,n }=\int _{ 0 }^{ \pi /2 }{ { sin }^{ m } } x{ cos }^{ n }xdx=\frac { n-1 }{ m+n } { I }_{ m,m-2 }n\ge 2\)
\(=\left( \frac { m-1 }{ n+m } \right) \left( \frac { m-3 }{ n+m-2 } \right) \left( \frac { m-5 }{ m+m-4 } \right) ...\frac { 2 }{ n+3 } .\frac { 1 }{ n+1 } \)
Here m = 2, n = 4
\(\therefore I=\frac { 3 }{ 6 } \times \frac { 1 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 32 } \)

Put x = sin \(\theta\). Then, dx = cos \(\theta\) d \(\theta\)
when x = 0, sin \(\theta\) = 0 and so \(\theta\) = 0. When x = 1, sin \(\theta\) = 1 and so \(\theta =\frac{\pi}{2}\)
Hence, we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { (1-{ sin }^{ 2 }\theta ) }^{ 5 }cos\theta d\theta } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }\theta { cos }^{ 11 }\theta d\theta =\frac { 10 }{ 16 } \times \frac { 8 }{ 14 } \times \frac { 6 }{ 12 } \times \frac { 4 }{ 10 } \times \frac { 2 }{ 8 } \times \frac { 1 }{ 6 } =\frac { 1 }{ 336 } } \)
By applying the reduction formula III iteratively, we get the following results (stated without proof):
\(\int _{ 0 }^{ 1 }{ { x }^{ m }{ (1-x) }^{ n }dx } =\frac { m!\times n! }{ (m+n+1)! } \), where m and n are positive integers

I =  \(\int ^\frac{\pi}{2} _0\)(3cos ⁴x-7sin⁵x)dx = 3\(\int ^\frac{\pi}{2} _0\)cos⁴ x dx-7\(\int ^\frac{\pi}{2} _0\)sin⁵ x dx
= 3 × \(\frac {3}{4}\) ×  \(\frac {1}{2}\) ×  \(\frac {\pi}{2}\) - 7 ×  \(\frac {4}{3}\) ×  \(\frac {2}{3}\)= \(\frac {9\pi}{16}\) - \(\frac {56}{15}\).
By applying the reduction formula III iteratively, we get the following results (stated without proof):
(i) If n is even and m is even,
\(\int ^\frac{\pi}{2}_{0}\)  sin ^m x cos ⁿ x dx = \(\frac {(n-1)}{m+n}\)  \(\frac {(n-3)}{m+n-2}\) \(\frac {(n-5)}{m+n-4}\) ......\(\frac {1}{m+2}\) \(\frac {m-1}{m}\) \(\frac {m-3}{m-2}\)\(\frac {m-5}{m-4}\) ... \(\frac {1}{2}\)\(\frac {\pi}{2}\)
(ii) If n is odd and m is any positive integer (even or odd), then
\(\int ^\frac{\pi}{2}_{0}\)  sin ^m x cosⁿx dx = \(\frac {(n-1)}{m+n}\) \(\frac {(n-3)}{m+n-2}\) \(\frac {(n-5)}{m+n-4}\) ... \(\frac {2}{m+3}\)\(\frac {1}{m+1}\)

\(\int _{ b }^{ \infty }{ \frac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } dx } ={ \left[ \frac { 1 }{ a } { tan }^{ -1 }\frac { x }{ a } \right] }_{ b }^{ \infty }=\frac { 1 }{ a } { tan }^{ -1 }\infty -\frac { 1 }{ a } { tan }^{ -1 }\frac { b }{ a } =\frac { 1 }{ a } \left[ \frac { \pi }{ 2 } -{ tan }^{ -1 }\frac { b }{ a } \right] \)

\(I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { 1 }{ 1+\sqrt { tanx } } dx } \int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { 1 }{ 1+\frac { \sqrt { sin\quad x } }{ \sqrt { cos\quad x } } } dx } \)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx\quad \quad ..(1) } \)
By the property,\( \int _{ a }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(a+b-x)dx } } \)
we get \(I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } }{ \sqrt { cos(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } +\sqrt { sin(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } } } dx\)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos(\frac { \pi }{ 2 } -x) } }{ \sqrt { cos(\frac { \pi }{ 2 } -x) } +\sqrt { sin(\frac { \pi }{ 2 } -x) } } } dx\)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { sin\quad x } }{ \sqrt { sin\quad x } +\sqrt { cos\quad x } } ...(2) } \)
\((1)+(2)\rightarrow \)
\(2I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx+\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { sin\quad x } }{ \sqrt { sin\quad x } +\sqrt { sin\quad x } } } } \)
\(=\int _{ \frac { \pi }{ 8 } }^{ 3\frac { \pi }{ 8 } }{ \frac { \sqrt { cos\quad x } +\sqrt { sin\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ dx={ [x] }_{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } } } } \)
\(2I=\frac { 3\pi }{ 8 } -\frac { \pi }{ 8 } =\frac { 2\pi }{ 8 } =\frac { \pi }{ 4 } \)
\(\therefore I=\frac { \pi }{ 8 } \)