Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int x \cos ^{2} x d x$

Solution:

Let $I=\int x \cos ^{2} x d x$

Using integration by parts,

$I=x \int \cos ^{2} x d x-\int \frac{d}{d x} x \int \cos ^{2} x d x$

We know that, $\cos ^{2} x=\frac{\cos 2 x+1}{2}$

$=x \int\left[\frac{\cos 2 x+1}{2}\right] d x-\int\left[1 \int\left[\frac{\cos 2 x+1}{2}\right] d x\right] d x$

We know that,

$\int \cos n x=\frac{\sin n x}{n}$

$=\frac{x}{2}\left[\frac{\sin 2 x}{2}+x\right]-\frac{1}{2} \int\left(x+\frac{\sin 2 x}{2}\right) d x$

$=\frac{x}{4} \sin 2 x+\frac{x^{2}}{2}-\frac{1}{2} \times \frac{x^{2}}{2}-\frac{1}{4}\left(-\frac{\cos 2 x}{2}\right)+c$

$I=\frac{x}{4} \sin 2 x+\frac{x^{2}}{4}+\frac{1}{8} \cos 2 x+c$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now