: Let I = ( - 1 x 4 ) 1 + x 8 dx Put - 1 x 4 = t 1 + x 8 4 x 3 dx = dt I = 4 = 4 ( - t ) + C = - 4 ( - 1 x 4 ) + C Choose the correct answer in Exercises 38 and 39.: ;

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.37,Page 305

Question

$\cfrac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}$

Step-by-step Solution

: Let $I = \int {\cfrac{{{x^3}\sin \left( {{{\tan }^{ - 1}}{x^4}} \right)}}{{1 + {x^8}}}} dx$

Put ${\tan ^{ - 1}}{x^4} = t$
$\Rightarrow$ $\cfrac{1}{{1 + {x^8}}} \cdot 4{x^3}dx = dt$

$\therefore$ $I = \cfrac{1}{4}\int {\sin tdt} = \cfrac{1}{4}\left( { - \cos t} \right) + C = - \cfrac{1}{4}\cos \left( {{{\tan }^{ - 1}}{x^4}} \right) + C$

$figure$

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