Area under the curve y = sin x between x = 0 and x = \(\frac { \pi }{ 3 } \) is
\({ A }_{ 1 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ ydx } =\int _{ 0 }^{ \frac { \pi }{ 3 } }{ sinxdx } =-{ \left[ cosx \right] }_{ 0 }^{ \frac { \pi }{ 3 } }\)
\(=-(cos\frac { \pi }{ 3 } -cos0)=-\left( \frac { 1 }{ 2 } -1 \right) \)
\(=-\left( -\frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } \)
Area under the curve y = sin 2x between x = 0 and \(\frac { \pi }{ 3 } \) is
\({ A }_{ 2 }=\int _{ 0 }^{ \frac { \pi }{ 3 } }{ { sin2 \ x \ dx=-\left[ \frac { cos2 \ x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 3 } } } \)
\(=-\frac { 1 }{ 2 } [cos2\frac { \pi }{ 3 } -cos0]-\frac { 1 }{ 2 } \left[ -\frac { 1 }{ 2 } -1 \right] =-\frac { 1 }{ 2 } \left( -\frac { 3 }{ 2 } \right) =\frac { 3 }{ 4 } \)
\(\therefore \frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { \frac { 1 }{ 2 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 2 } \times \frac { 4 }{ 3 } =\frac { 2 }{ 3 } \)
∴ A₁:A₂ =2 : 3 

We note that \(y=\frac { 3 }{ 4 } \sqrt { { x }^{ 2 }-16 } \Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1\). So, the given curve is a portion of the hyperbola \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1\) between the lines y = 1 and y = 6 and it lies above the x-axis.

The region to be revolved is sketched.
Since revolution is made about y-axis, we write the equation of the portion of the hyperbola as \(x=\frac { 4 }{ 3 } \sqrt { 9+{ y }^{ 2 } } .\) 
So, the volume of the solid generated is given by 
\(V=\pi \int _{ 1 }^{ 6 }{ { x }^{ 2 }dy=\pi \int _{ 1 }^{ 6 }{ { \left( \frac { 4 }{ 3 } \sqrt { 9+{ y }^{ 2 } } \right) }^{ 2 } } } dy=\pi \left( \frac { 16 }{ 9 } \right) \int _{ 1 }^{ 16 }{ \left( 9+{ y }^{ 2 } \right) dy } \)
\(=\pi \left( \frac { 16 }{ 9 } \right) { \left( 9y+\frac { { y }^{ 3 } }{ 3 } \right) }_{ 1 }^{ 6 }=\pi \left( \frac { 16 }{ 9 } \right) \left[ (54+72)-(9+\frac { 1 }{ 3 } ) \right] =\frac { 5600 }{ 27 } \pi \)

I =\(\int ^{1}_{0} (tan ^{-1} x + tan ^{-1}(1-x))\) dx
=\(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-x)) dx\)
= \(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-(1-x)) dx\),  Since\(\int ^{a}_{0} f (x) dx = \int ^{a}_{0} f(a-x) dx \)
=\(\int ^{1}_{0} tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1} x dx\)
= 2\(\int ^{1}_{0} tan^{-1} x dx\)
= \([2 \int udv]^1_0\) ,  Where u = tan^-1 x and dν = dx
= 2\([uν - \int udv]^1_0,\) applying integration by parts
= 2\((x tan^{-1} x - \int x \frac{dx}{1+x^2})^{1}_{0}\)
= 2\((x tan^{-1} x - \frac {1}{2} log (1+x^2))^{1}_{0}\)
= \(\frac{\pi}{2} \)- log 2

Put u = tan \(\frac{x}{2}\) 
Then, sin x = \(\frac{2 tan \frac {x}{2}}{1 + tan^2 \frac{x}{2}}\) = \(\frac{2u}{1+u^2}\), du = \(\frac{1}{2}\) sec² \(\frac{x}{2}\) dx  \(\Rightarrow \) dx = \(\frac {2du}{1+u^2}\)
When x = 0,u = tan 0 = 0
When x = \(\frac{\pi}{2},u=tan\frac{\pi}{4}=1\)
∴ I =\(\int ^\frac{\pi}{2}_0\) \(\frac{dx}{4+5 sin x}\) =\(\int ^{1}_0\) \(\frac{\frac {2du}{1+u^2}}{4+5 (\frac{2u}{1+u^2})}\) =\(\int ^{1}_0\) \(\frac {du}{2u^2+5u +2}\)=\(\frac{1}{2}\)\(\int ^{1}_{0}\) \(\frac {du}{u^2+\frac{5}{2}u +1}\)
\(\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { du }{ (u+\frac { 5 }{ 4 } )^2-(\frac { 3 }{ 4 } )^{ 2 } } { \left[ \frac { 1 }{ 2 } \times \frac { 1 }{ 2\times \left( \frac { 3 }{ 4 } \right) } log\left( \frac { (u+\frac { 5 }{ 4 } )-\frac { 3 }{ 4 } }{ (u+\frac { 5 }{ 4 } )+\frac { 3 }{ 4 } } \right) \right] }_{ 0 }^{ 1 } = \frac{1}{3} [log (\frac{u+\frac{1}{2}}{u+2})] =\frac { 1 }{ 3 } log2\)

Let I = \(\int _{ 1 }^{ 2 }{ \frac { x }{ (x+1)(x+2) } dx } \)
\(I=\int _{ 1 }^{ 2 }{ \left[ \frac { -1 }{ (x+1) } +\frac { 2 }{ x+2 } \right] } dx\) (Using partial fractions)
\(={ [-log(x+1)+2log(x+2)] }_{ 1 }^{ 2 }\)
\(=log{ \left[ \frac { { (x+2) }^{ 2 } }{ x+1 } \right] }_{ 1 }^{ 2 }\)
\(=log\frac { 16 }{ 3 } -log\frac { 9 }{ 2 } \)
\(=log\frac { 32 }{ 27 } \)

Given curve is y = \(\sqrt x\) + 1

Required area
\(\int _{ 0 }^{ 4 }{ ydx } =\int _{ 0 }^{ 4 }{ (\sqrt { x } +1)dx } \)
\({ \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+x \right] }_{ 0 }^{ 4 }=\frac { 2 }{ 3 } { (4) }^{ \frac { 3 }{ 2 } }+4\)
\(\frac { 2 }{ 3 } (4)\sqrt { 4 } +4=\frac { 16 }{ 3 } +4\)
\(\frac { 16+12 }{ 3 } =\frac { 28 }{ 3 } \) sq.units

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

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 \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sin \ x }{ 9+{ cos }^{ 2 } } dx } \)

Put t = cos x ⇒ - sin x dx = dt ⇒ sin x dx = - dt
\(\therefore I=\int _{ 1 }^{ 0 }{ -\frac { dt }{ { 3 }^{ 2 }+t^{ 2 } } } =\int _{ 0 }^{ 1 }{ \frac { dt }{ { 3 }^{ 2 }+t^{ 2 } } } =\frac { 1 }{ 3 } { \left[ { tan }^{ -1 }\left( \frac { t }{ 3 } \right) \right] }_{ 0 }^{ 1 }\)
= \(\frac { 1 }{ 3 } \left[ { tan }^{ -1 }\left( \frac { 1 }{ 3 } \right) -{ tan }^{ -1 }(0) \right] \)
= \(\frac { 1 }{ 3 } { tan }^{ -1 }\left( \frac { 1 }{ 3 } \right) \) [∵ tan^-1(0) = 0]

Using the formula \(n={ e }^{ { log }_{ e }n },\) we get
\(I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { n }^{ x } } } dx=\int _{ 0 }^{ \infty }{ { n }^{ -x }{ x }^{ n }dx=\int _{ 0 }^{ \infty }{ { n }^{ -x }{ x }^{ n } } dx=\int _{ 0 }^{ \infty }{ { ({ e }^{ log\quad n }) }^{ -x }{ x }^{ n }dx=\int _{ 0 }^{ \infty }{ { e }^{ -xlog\quad n }{ x }^{ n }dx } } } \)
Using the substitution u = x log n, we get dx \(\frac { du }{ log\ n } \)
When x = 0,
we get u = 0
When x = \(\infty\), we get u = \(\infty\)
\(\therefore I=\int _{ 0 }^{ \infty }{ { e }^{ -u }{ \left( \frac { u }{ log\quad n } \right) }^{ n }\frac { du }{ log\quad n } } \)
\(=\frac { 1 }{ { (log\quad n) }^{ n+1 } } \int _{ 0 }^{ \infty }{ { e }^{ -u }{ u }^{ (n+1)-1 }du=\frac { Γ(n+1) }{ { (log\quad n) }^{ n+1 } } =\frac { n! }{ { (log) }^{ n+1 } } } \)

\(\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } } \)
\(=\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+1+4 } } =\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| \right] \)
\(={ \left[ \frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { x+1 }{ 2 } \right) \right] }_{ -1 }^{ 1 }\)
\(=\frac { 1 }{ 2 } \left[ { tan }^{ -1 }(1)-{ tan }^{ -1 }(0) \right] \)
\(\\ =\frac { 1 }{ 2 } \left[ \frac { \pi }{ 4 } \right] =\frac { \pi }{ 8 } \)

(4) \(\int _{ 0 }^{ a }{ f(a-x) } dx\)

(3) \(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ (sinx-cosx) } dx\)