#### need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 39

Answer: $\frac{1}{3} \sec ^{3}\left(x^{2}+3\right)+c$

Hint: Use substitution method to solve this integral.

Given:   $\int 2 x \cdot \sec ^{3}\left(x^{2}+3\right) \tan \left(x^{2}+3\right) d x$

Solution:

\begin{aligned} &\text { Let } I=\int 2 x \cdot \sec ^{3}\left(x^{2}+3\right) \tan \left(x^{2}+3\right) d x \\ &\text { Put } x^{2}+3=t \Rightarrow 2 x d x=d t \\ &\Rightarrow d x=\frac{d t}{2 x} \text { then } \end{aligned}

\begin{aligned} \Rightarrow I &=\int 2 x \sec ^{3} t \cdot \tan t \frac{d t}{2 x}=\int \sec ^{3}(t) \cdot \tan (t) d t \\ &=\int \sec ^{2}(t) \sec (t) \tan (t) d t \end{aligned}

$\text { Again Put } \sec t=u \Rightarrow \sec t \tan t \; d t=d u \text { then }$

$I=\int u^{2} d u=\left[\frac{u^{2+1}}{2+1}\right]+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

$\begin{array}{ll} =\frac{u^{3}}{3}+c=\frac{\sec ^{3} t}{3}+c & {[\because \sec t=u]} \end{array}$

$=\frac{1}{3} \sec ^{3}\left(x^{2}+3\right)+c \quad\left[\because t=x^{2}+3\right]$