\( \int\displaystyle\frac{\tan^5\sqrt{x}.\sec^2\sqrt{x}}{\sqrt{x}}dx \) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{1}{3}\tan^6\sqrt{x}+\mathrm{C} \) , where C is an arbitrary constant
Let \( t=\tan\sqrt{x} \). By the chain rule:
\( \displaystyle dt=\sec^2\sqrt{x}\cdot\frac{1}{2\sqrt{x}}\,dx \)
\( \displaystyle \frac{\sec^2\sqrt{x}}{\sqrt{x}}\,dx=2\,dt \)
So
\( \displaystyle \int\frac{\tan^5\sqrt{x}\,\sec^2\sqrt{x}}{\sqrt{x}}\,dx=\int 2t^5\,dt \)
\( \displaystyle =\frac{2t^6}{6}+C \)
\( \displaystyle =\frac{1}{3}\tan^6\sqrt{x}+C \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{3}\tan^5\sqrt{x}+\mathrm{C} \) , where C is an arbitrary constant — Incorrect. The power of tan must rise from 5 to 6 on integrating; this keeps the old power.
Option (3) \( \displaystyle\frac{1}{5}\tan^5\sqrt{x}+\mathrm{C} \) , where C is an arbitrary constant — Incorrect. This keeps the power 5 and divides by 5, which is not the power rule \( \int t^n\,dt=\frac{t^{n+1}}{n+1} \).
Option (4) \( \tan^6\sqrt{x}+\mathrm{C} \) , where C is an arbitrary constant — Incorrect. It misses the factor \( \frac{2}{6}=\frac{1}{3} \); its derivative is 3 times the integrand.