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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
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.} $

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