Evaluate the integral:

Question:

Evaluate the integral:

$\int \cos x \sqrt{4-\sin ^{2} x} d x$

Solution:

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. $\mathrm{t}=\sin \mathrm{x}$

$\Rightarrow I=\frac{1}{2} \sin x \sqrt{4-\sin ^{2} x}+2 \sin ^{-1}\left(\frac{\sin x}{2}\right)+C$