{x³ (^{-1} x^4)}{1+x^8} - NCERT Class XII Mathematics Chapter

NCERT Class XII Mathematics Chapter - - Solutions

Exercise - 7.2
Integrate the functions in Exercises 23 to 59.

Question : 59 of 271

Marks: +1, -0

\frac{x^3 \sin(\tan^{-1} x^4)}{1+x^8}

Solution:

Let

I = \int \frac{x^3 \sin(\tan^{-1} x^4)}{1+x^8} \, dx

Put

\tan^{-1} x^4 = t

\Rightarrow \frac{1}{1+x^8} \cdot 4x^3 \, dx = dt

\therefore I = \frac{1}{4} \int \sin t \, dt

= \frac{1}{4} (-\cos t) + C

= -\frac{1}{4} \cos(\tan^{-1} x^4) + C

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