The given curve is \(y=\begin{cases} cosx,0\le x\le \frac { \pi }{ 2 } \\ -cosx,\frac { \pi }{ 2 } \le x\le \pi \end{cases}\)
It lies above the x − axis. The required area is sketched. So, the required area is given by

\(A=\int _{ 0 }^{ \pi }{ ydx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cosxdx } +\int _{ \frac { \pi }{ 2 } }^{ \pi }{ (-cosx)dx } ={ [sin\quad x] }_{ 0 }^{ \frac { \pi }{ 2 } }-{ [sin\quad x] }_{ \frac { \pi }{ 2 } }^{ \pi } } \)
= [1-0]-[0-1] = 2

Let \(I=\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx } \quad ...(1)\)
By the property \(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a-x)dx } } \right] \)
we get 
\(I=\int _{ 0 }^{ 2\pi }{ { \left[ sin(2\pi -x) \right] }^{ 4 } } { \left[ cos(2\pi -x) \right] }^{ 3 }dx\)
\(=\int _{ 0 }^{ 2\pi }{ { [-sin\quad x] }^{ 4 }{ [-cos\quad x] }^{ 3 }dx } \)
\(=-\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx\quad ..(2) } \)
\(\therefore (1)+(2)\Rightarrow \)
\(2I=\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx } \)
\(-\int _{ 0 }^{ 2\pi }{ { sin }^{ 4 }x{ cos }^{ 3 }xdx=0 } \)
\(\Rightarrow I=0\)

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

Let I = \(\int _{ 0 }^{ 1 }{ \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } dx } \)

Put e^x = t ⇒ e^x dx = dt
∴ \(\int _{ 1 }^{ e }{ \frac { dt }{ 1+{ t }^{ 2 } } dx } { \left[ { tan }^{ -1 }(t) \right] }_{ 1 }^{ e }\)
= tan^-1(e) - tan^-1(1)
= tan^-1(e) -\(\frac { \pi }{ 4 } \)

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\(\frac {1}{2} \times \frac {\pi}{2} + \frac {3}{4} \times \frac {1}{2} \times \frac {\pi}{2} = \frac {7\pi}{16} \)

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 = \(\int _{ 0 }^{ 1 }{ \sqrt { 9-4{ x }^{ 2 } } dx } =\int _{ 0 }^{ 1 }{ 2\sqrt { \left( \frac { 3 }{ 2 } \right) ^{ 2 }-{ x }^{ 2 } } } dx\)
\(={ 2\left[ \frac { x }{ 2 } \sqrt { \left( \frac { 3 }{ 2 } \right) ^{ 2 }-{ x }^{ 2 } } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { x }{ \frac { 3 }{ 2 } } \right) \right] }_{ 0 }^{ 1 }\)
\(\left[ \because \sqrt { { a }^{ 2 }-{ x }^{ 2 } } =\frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } \left( \frac { x }{ a } \right) \right] \)
\(=2\left[ \frac { 1 }{ 2 } \sqrt { \frac { 9 }{ 4 } -1 } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 2 }{ 3 } \right) -0 \right] \)
\(=2\left[ \frac { 1 }{ 2 } \sqrt { \frac { 5 }{ 4 } } +\frac { 9 }{ 8 } { sin }^{ -1 }\left( \frac { 2 }{ 3 } \right) \right] \)
\(=\frac { \sqrt { 5 } }{ 2 } +\frac { 9 }{ 4 } { sin }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)

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]

The parabola x = y² +1 is y² x = −1. It is symmetrical about x-axis and has the vertex at (1, 0) and focus at \(\left( \frac { 5 }{ 4 } ,0 \right) \). The region for revolution is shaded. Hence, the required volume is given by
\(V=\pi \int _{ -1 }^{ 1 }{ { x }^{ 2 }dy } \)
\(=\pi \int _{ -1 }^{ 1 }{ { ({ y }^{ 2 }+1) }^{ 2 }dy } \)
\(=2\pi \int _{ 0 }^{ 1 }{ \left( { y }^{ 4 }+{ 1y }^{ 2 }+1 \right) dy } \), since the integrand is an even function
\(=2\pi { \left( \frac { { y }^{ 5 } }{ 5 } +2\frac { { y }^{ 3 } }{ 3 } +y \right) }_{ 0 }^{ 1 }=2\pi \left( \frac { 1 }{ 5 } +\frac { 2 }{ 3 } +1 \right) \pi \)

Let \(u={ x }^{ 3 }\quad dv={ e }^{ -2x }\)
\(u_1=3{ x }^{ 2 }{ v }_{ 1 }=\frac { { e }^{ -2x } }{ -2 } \)
\(u_2=6x\quad { v }_{ 2 }=\frac { { e }^{ -2x } }{ 4 } \)
\(u_3=6\quad { v }_{ 3 }=\frac { { e }^{ -2x } }{ -8 } \)
\(u_4 = 0 \quad \ { v }_{ 4 }=\frac { { e }^{ -2x } }{ 16 } \)
Bernoulli's' formula:
\(\int { uvdx } ={ uv }_{ 1 }-u'{ v }_{ 2 }+u"{ v }_{ 3 }-u"'{ v }_{ 4 }+...\)
\(\therefore \int _{ 0 }^{ 1 }{ { x }^{ 3 }{ e }^{ -2x }dx={ \left[ { x }^{ 3 }\left( \frac { { e }^{ -2x } }{ -2 } \right) -3{ x }^{ 2 }\left( \frac { { e }^{ -2x } }{ 4 } \right) +6x\left( \frac { { e }^{ -2x } }{ -8 } \right) -6\left( \frac { { e }^{ -2x } }{ 16 } \right) \right] }_{ 0 }^{ 1 } } \)
\(={ \left[ { e }^{ -2x }\left( \frac { -{ x }^{ 3 } }{ 2 } -\frac { -3{ x }^{ 2 } }{ 4 } -\frac { 3x }{ 4 } -\frac { 3 }{ 8 } \right) \right] }_{ 0 }^{ 1 }\)
\(=\left[ { e }^{ -2 }\left( -\frac { 1 }{ 2 } -\frac { 3 }{ 4 } -\frac { 3 }{ 4 } -\frac { 3 }{ 8 } \right) -{ e }^{ -0 }\left( \frac { -3 }{ 8 } \right) \right] \)
\(={ e }^{ -2 }\left( \frac { -4-12-3 }{ 8 } \right) +\frac { 3 }{ 8 } \)
\(={ e }^{ -2 }\left( \frac { -19 }{ 8 } \right) +\frac { 3 }{ 8 } =\frac { 3 }{ 8 } -\frac { 19 }{ 8 } { e }^{ -2 }\)

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