MCQ
$\int {\frac{{{{\sin }^3}2x}}{{{{\cos }^5}2x}}dx = } $
- A${\tan ^4}x + C$
- B$\tan 4x + C$
- C${\tan ^4}2x + x + C$
- ✓$\frac{1}{8}{\tan ^4}2x + C$
✓
Answer
Correct option: D.
$\frac{1}{8}{\tan ^4}2x + C$
d
(d) $I = \int {\frac{{{{\sin }^3}2x}}{{{{\cos }^5}2x}}dx} $
==> $I = \int {\frac{{{{\sin }^3}2x}}{{{{\cos }^3}2x}}.\frac{1}{{{{\cos }^2}2x}}dx = \int {{{\tan }^3}2x.{{\sec }^2}2x\,dx.} } $
Putting tan $2x = t$ and $2{\sec ^2}2x\,dx = dt$, we get
$I = \int {{t^3}\frac{{dt}}{2} = \frac{1}{2}.\frac{{{t^4}}}{4} + C = \frac{1}{8}({{\tan }^4}2x) + C.} $
(d) $I = \int {\frac{{{{\sin }^3}2x}}{{{{\cos }^5}2x}}dx} $
==> $I = \int {\frac{{{{\sin }^3}2x}}{{{{\cos }^3}2x}}.\frac{1}{{{{\cos }^2}2x}}dx = \int {{{\tan }^3}2x.{{\sec }^2}2x\,dx.} } $
Putting tan $2x = t$ and $2{\sec ^2}2x\,dx = dt$, we get
$I = \int {{t^3}\frac{{dt}}{2} = \frac{1}{2}.\frac{{{t^4}}}{4} + C = \frac{1}{8}({{\tan }^4}2x) + C.} $
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