Evaluate the integral: \(\int { {x^2\over1+x^3}dx. } \)

1

(a)

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

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: \(\int {sec^2\ x \over cosec^2 x}dx.\)

1

(a)

\(\int \frac{\sin ^{2} x}{\cos ^{2} x} d x=\int \tan ^{2} x d x=\int\left(\sec ^{2} x-1\right) d x=\tan x-x+C\)

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

1

(a)

= 2 tan x - 3 sec x + c

\(\int {1+tan\ x\over 1-tan\ x}dx\).

1

(a)

\(\int \frac{1+\tan x}{1-\tan x} d x =\int \frac{\cos x+\sin x}{\cos x-\sin x} d x \)
\(=-\int \frac{1}{t} d t \)
\(=-\log |t|+C \)
\(=-\log |\cos x-\sin x|+C\)

\(\int {dx\over x\ cos^2 (1+log\ x)}\).

1

(a)

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

\(\int cos^2x\ cosec^2x\ dx.\)

1

(a)

\(= -{1\over3}cot^3\ x+c\)

\(\int {x^2-1\over x^2+1}dx.\)

1

(a)

= x-2 tan^-1x + c

\(\int tan^{-1}(cot\ x)dx.\)

1

(a)

\(\text { We have }\int \tan ^{-1}(\cot x) d x=\int \tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x=\int\left(\frac{\pi}{2}-x\right) d x=\frac{\pi}{2} x-\frac{x^{2}}{2}+C \)

\(\int \sqrt{tan\ x}(1+tan^2\ x)\ dx.\)

1

(a)

\(\int \sqrt{\tan x} \cdot \sec ^{2} x d x=\int \sqrt{t} d t=\frac{2}{3} t^{\frac{3}{2}}+C=\frac{2}{3}(\tan x)^{\frac{3}{2}}+C\)