MCQ
$\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}=$
  • A
    $\frac{1}{3}\tan^2\text{x}+\text{C}$
  • B
    $\frac{1}{2}\tan^2\text{x}+\text{C}$
  • $\frac{1}{3}\tan^3\text{x}+\text{C}$
  • D
    none of these.

Answer

Correct option: C.
$\frac{1}{3}\tan^3\text{x}+\text{C}$
$\text{I}=\int\frac{\sin^2\text{x}}{\cos^4\text{x}}\text{ dx}$
$\text{I}=\int\tan^{2}\text{x}\sec^2\text{x dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\text{t}^2\text{dt}$
$\text{I}=\frac{\text{t}^3}{3}+\text{C}$
$\text{I}=\frac{\tan^3\text{x}}{3}+\text{C}$

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