Question
Integrate the following functions w.r.t. x:
$\frac{x \cdot \sec ^2\left(x^2\right)}{\sqrt{\tan ^3\left(x^2\right)}}$
$\frac{x \cdot \sec ^2\left(x^2\right)}{\sqrt{\tan ^3\left(x^2\right)}}$
✓
Answer
Let $I=\int \frac{x \cdot \sec ^2\left(x^2\right)}{\sqrt{\tan ^3\left(x^2\right)}} d x$
Put $\tan \left(x^2\right)=t \quad \therefore \sec ^2\left(x^2\right) \times 2 x d x=d t$
$\therefore x \cdot \sec ^2\left(x^2\right) d x=\frac{d t}{2}$
$\therefore I=\int \frac{1}{\sqrt{t^3}} \cdot \frac{d t}{2}=\frac{1}{2} \int t^{-\frac{3}{2}} d t$
$=\frac{1}{2} \cdot \frac{t^{-\frac{1}{2}}}{-1 / 2}+c=\frac{-1}{\sqrt{t}}+c=\frac{-1}{\sqrt{\tan \left(x^2\right)}}+c$.
Put $\tan \left(x^2\right)=t \quad \therefore \sec ^2\left(x^2\right) \times 2 x d x=d t$
$\therefore x \cdot \sec ^2\left(x^2\right) d x=\frac{d t}{2}$
$\therefore I=\int \frac{1}{\sqrt{t^3}} \cdot \frac{d t}{2}=\frac{1}{2} \int t^{-\frac{3}{2}} d t$
$=\frac{1}{2} \cdot \frac{t^{-\frac{1}{2}}}{-1 / 2}+c=\frac{-1}{\sqrt{t}}+c=\frac{-1}{\sqrt{\tan \left(x^2\right)}}+c$.
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