Evaluate the integral:
$\int \cos x \sqrt{4-\sin ^{2} x} d x$
Key points to solve the problem:
- Such problems require the use of the method of substitution along with a method of integration by parts. By the method of integration by parts if we have $\int \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx}=\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x}) \mathrm{dx}-\int \mathrm{f}^{\prime}(\mathrm{x})\left(\int \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}\right) \mathrm{dx}$
- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$
$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$
$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$
Let, $I=\int \cos x \sqrt{4-\sin ^{2} x} d x$
Let, $\sin x=t$
Differentiating both sides:
$\Rightarrow \cos x d x=d t$
Substituting $\sin x$ with $t$, we have:
$\therefore I=\int \sqrt{4-\mathrm{t}^{2}} \mathrm{dt}=\int \sqrt{2^{2}-\mathrm{t}^{2}} \mathrm{dt}$
As I match with the form: $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$
$\therefore \mathrm{I}=\frac{\mathrm{t}}{2} \sqrt{4-(\mathrm{t})^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{\mathrm{t}}{2}\right)+\mathrm{C}$
Putting the value of $\mathrm{t}$ i.e. $\mathrm{t}=\sin \mathrm{x}$
$\Rightarrow I=\frac{1}{2} \sin x \sqrt{4-\sin ^{2} x}+2 \sin ^{-1}\left(\frac{\sin x}{2}\right)+C$
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.