CUET Preparation Today
\(\int \frac{\tan^{4}\sqrt{x} \sec^{2}\sqrt{x}dx}{\sqrt{x}}\) equals |
\(\frac{2}{5}\tan^{2}\sqrt{x}+C\) \(\frac{2}{5}\tan^{5}\sqrt{x}+C\) \(\frac{5}{2}\tan^{5}\sqrt{x}+C\) None |
\(\frac{2}{5}\tan^{5}\sqrt{x}+C\) |
\(I=\int \frac{\tan^{4}\sqrt{x} \sec^{2}\sqrt{x}}{\sqrt{x}}dx\) let $y=\tan\sqrt{x}$ $2dy=\frac{\sec^2\sqrt{x}}{\sqrt{x}}dx$ so $I=2\int y^4dy=\frac{2}{5}y^5+C$ $=\frac{2}{5}\tan^{5}\sqrt{x}+C$ |
