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JEE Mathematics Solved Question Paper 2008

Multiple Choice Questions

21.
$\int \frac{\sin (x) \cos (x)}{\sqrt{1 - \sin^{4} (x)}} dx$ is equal to

$\frac{1}{2} \sin^{- 1} (\sin^{2} (x)) + C$

$\frac{1}{2} \cos^{- 1} (\sin^{2} (x)) + C$

$\tan^{- 1} (\sin^{2} (x)) + C$

$\tan^{- 1} (2 \sin^{2} (x)) + C$

$\frac{1}{2} \sin^{- 1} (\sin^{2} (x)) + C$

$Let I = \int \frac{\sin (x) \cos (x)}{\sqrt{1 - \sin^{4} (x)}} dx Put \sin^{2} (x) = t \Rightarrow 2 \sin (x) \cos (x) dx = dt ∴ I = \int \frac{dt}{2 \sqrt{1 - t^{2}}} = \frac{1}{2} \sin^{- 1} (t) + C = \frac{1}{2} \sin^{- 1} (\sin^{2} (x)) + C$

NCERT Solutions
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Class 8 NCERT Maths Solutions

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Multiple Choice Questions

21. ∫sinxcosx1 - sin4xdx is equal to 12sin-1sin2x + C 12cos-1sin2x + C tan-1sin2x + C tan-12sin2x + C

NCERT SolutionsTextbook Solutions | Additional Questions

21.
$\int \frac{\sin (x) \cos (x)}{\sqrt{1 - \sin^{4} (x)}} dx$ is equal to

$\frac{1}{2} \sin^{- 1} (\sin^{2} (x)) + C$

$\frac{1}{2} \cos^{- 1} (\sin^{2} (x)) + C$

$\tan^{- 1} (\sin^{2} (x)) + C$

$\tan^{- 1} (2 \sin^{2} (x)) + C$

NCERT Solutions
Textbook Solutions | Additional Questions