COMPASSLearning

Talk to us Register Login

CBSE Class 12 · 2007 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{c o s x}{( 1 - s i n x ) ( 2 - s i n x )} d x

Answer: $lo g \frac{2 - sin x}{1 - sin x} + C$

Solution

Put

t = sin x

,

d t = cos x d x

:

\int \frac{d t}{( 1 - t ) ( 2 - t )} .

Partial fractions:

\frac{1}{( 1 - t ) ( 2 - t )} = \frac{1}{1 - t} - \frac{1}{2 - t}

, so

\int [\frac{1}{1 - t} - \frac{1}{2 - t}] d t = - lo g ∣1 - t ∣ + lo g ∣2 - t ∣ + C = lo g \frac{2 - t}{1 - t} + C .

Restoring

t = sin x

,

lo g \frac{2 - sin x}{1 - sin x} + C

CBSE Class 12 Board · 2007 · Subjective

Practise more: all CBSE board PYQs on Indefinite Integrals →

More board questions on Indefinite Integrals

The integral $\int \frac{d x}{9 - 4 x ^{2}}$ is equal to
CBSE 2024View solution →
Find $\int cos^{3} x e^{l o g s i n x} d x$
CBSE 2024View solution →
Find $\int \frac{2 x + 1}{( x + 1 ) ^{2} ( x - 1 )} d x$
CBSE 2024View solution →
Anti-derivative of $\frac{tan x - 1}{tan x + 1}$ with respect to $x$ is:
CBSE 2023View solution →
Find $\int \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} d x$
CBSE 2023View solution →
Evaluate $\int e^{x} (\frac{1 - s i n x}{1 - c o s x}) d x$
CBSE 2023View solution →