Evaluate the integral:

Question:

Evaluate the integral:

$\int \sqrt{3-x^{2}} d x$

Solution:

Key points to solve the problem:

- Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have $\int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x)\left(\int g(x) d x\right) d x$

- 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, $I=\int \sqrt{3-x^{2}} d x$

$\therefore I=\int \sqrt{3-\mathrm{x}^{2}} \mathrm{dx}=\int \sqrt{(\sqrt{3})^{2}-\mathrm{x}^{2}} \mathrm{dx}$

As I match with the form: $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$

$\therefore I=\frac{x}{2} \sqrt{3-x^{2}}+\frac{3}{2} \sin ^{-1}\left(\frac{x}{\sqrt{3}}\right)+C$

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