Evaluate the integral:  \(\int { {2cos\ x\over 3sin^2}dx. } \)

1

(a)

\(=-{2\over3}cosec\quad x+C\)

Evaluate the integral: \(\int { x\ +\ cos\ 6x \over 3x^2\ + sin\ 6x} dx.\)

1

(a)

\(\int \frac{x+\cos 6 x}{3 x^{2}+\sin 6 x} d x=\frac{1}{6} \int \frac{1}{t} d t=\frac{1}{6} \log |t|+C\)
\({1\over6}log|3x^2 + sin6x|+C|\)

Evaluate the integral: \(\int {sec^2x\over3+tan\ x}dx.\)

1

(a)

\(\int \frac{\sec ^{2} x}{3+\tan x} d x =\int \frac{1}{t} d t \)
\(=\log |t|+C \)
= log | 3 + tan x | + C

Evaluate the integral: \(\int {sec\ x\ (sec\ x\ +\ tan\ x)}dx\)

1

(a)

= tan x + sec x + c

Given \(\int {e^x (tan\ x\ +\ 1)\ sec\ x\ dx\ =\ e^x}\ f(x)\ +\ c.\) Write the value of f(x).

1

(a)

\(= log\ |{1-cos\ x\over 2-cos\ x}|+\ c\)

Evaluate the integral: \(\int({cos\ x-sin\ x})^2 dx\)

1

(a)

Consider : \(\int({cos\ x-sin\ x})^2dx=\int({cos^2x + sin^2\ x-2sin\ x\ \ cos\ x})dx\)
\(=\int({1-sin\ 2\ x})dx=x+{cos\ 2x\over 2}+ C\)

Evaluate the integral: \(\int sin^{-1}(cos\ x)dx.\)

1

(a)

\(= {\pi\over2}\times-{x^2\over 2}+c\)

Evaluate: \(\int { { cos }^{ - }(sinx)dx } \)

1

(a)

\(\int { { cos }^{ -1 }\left[ cos\left( \frac { \pi }{ 2 } -x \right) \right] dx } =\int { \left( \frac { \pi }{ 2 } -x \right) } dx\)
\(=\int { \frac { \pi }{ 2 } } dx-\int { xdx } \)
\(=\frac { \pi }{ 2 } x-\frac { { x }^{ 2 } }{ 2 } +c\)

Evaluate : \(\int _{ 1 }^{ 2 }{ \frac { { x }^{ 3 }-1 }{ { x }^{ 2 } } } dx\)

1

(a)

\(I=\int _{ 1 }^{ 2 }{ \left( \frac { { x }^{ 3 } }{ { x }^{ 2 } } -\frac { 1 }{ { x }^{ 2 } } \right) } dx\)
\(=\int _{ 1 }^{ 2 }{ \left( x-\frac { 1 }{ { x }^{ 2 } } \right) dx } \)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } +\frac { 1 }{ x } \right] ^{ 2 }_{ 1 }\)
\(I=\left( 2+\frac { 1 }{ 2 } \right) -\left( \frac { 1 }{ 2 } +1 \right) \)
\(=\frac { 5 }{ 2 } -\frac { 3 }{ 2 } \)
\(\Rightarrow I=1\)

Evaluate : \(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tanx } dx\)

1

(a)

\(\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tanx } dx=\left[ log\left| secx \right| \right] ^{ \pi /4 }_{ 0 }\)
\(=log\left| sec\frac { \pi }{ 4 } \right| -log\left| sec0 \right| \)
\(=log\left| \sqrt { 2 } \right| -log\left| 1 \right| \)
\(\therefore \ \int _{ 0 }^{ \pi /4 }{ tanxdx } =\frac { 1 }{ 2 } log2\)

Evaluate : \(\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } } \)

1

(a)

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

\(\int { sin2xcos3xdx } \)

2

(a)

\(\int { tan^{ -1 } } \sqrt { \frac { 1-cos2x }{ 1+cos2x } } dx\)

2

(a)

\(\int { \frac { (x+1){ e }^{ x } }{ cos^{ 2 }(xe^{ x }) } } dx\)

2

(a)

\(\int { { e }^{ x } } \left[ cotx+logsinx \right] dx\)

2

(a)

Evaluate : \(\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { \sqrt [ 3 ]{ sinx } }{ \sqrt [ 3 ]{ cosx } +\sqrt [ 3 ]{ sinx } } } dx\)

2

(a)

Evaluate : \(\int _{ 0 }^{ 1 }{ \frac { { e }^{ x } }{ 1+{ e }^{ 2x } } } dx.\)

3

(a)

If \(x=\int _{ 0 }^{ x }{ tan } \) sin t dt, then write the value of \(f^{ \prime }\left( x \right) \)

3

(a)

Find: \(\int { \frac { { x }^{ 2 }+x+1 }{ ({ x }+2)({ x }^{ 2 }+1) } } dx.\)

3

(a)

Find: \(\int { \log { x } dx } .\)

3

(a)

Find the integral: \(\int { (1-x) } \sqrt { x } dx\)

3

(a)

Evaluate the integral: \(\int {x^2+4\over x^4+16}dx.\)

4

(a)

Evaluate the integral: \(\int x^2\ tan^{-1}\ x\ dx\)

4

(a)

Find : \(\int { \frac { sin^{ -1 }\sqrt { x } -cos^{ -1 }\sqrt { x } }{ sin^{ -1 }\sqrt { x } +cos^{ -1 }\sqrt { x } } } dx,x\epsilon \left[ 0,1 \right] \)

6

(a)