CBSE Class 12 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{1}{c o s ^{4} x + s i n ^{4} x} d x

Answer: $\frac{1}{2} tan^{- 1} (\frac{tan x - cot x}{2}) + C$

Solution

Divide numerator and denominator by

cos^{4} x

\int \frac{sec ^{4} x}{1 + tan ^{4} x} d x = \int \frac{( 1 + tan ^{2} x ) sec ^{2} x}{1 + tan ^{4} x} d x .

Put

t = tan x

d t = sec^{2} x d x

\int \frac{1 + t ^{2}}{1 + t ^{4}} d t = \int \frac{1 + \frac{1}{t ^{2}}}{t ^{2} + \frac{1}{t ^{2}}} d t .

Now put

z = t - \frac{1}{t}

, so

d z = (1 + \frac{1}{t ^{2}}) d t

and

t^{2} + \frac{1}{t ^{2}} = z^{2} + 2

\int \frac{d z}{z ^{2} + 2} = \frac{1}{2} tan^{- 1} \frac{z}{2} + C .

With

z = tan x - cot x

\frac{1}{2} tan^{- 1} (\frac{tan x - cot x}{2}) + C

CBSE Class 12 Board · Subjective