7 x/ 9 x 2 x d x is equal to

Correct Answer is : 19log⁡sec⁡(9x)−12log⁡sec⁡(2x)+c1/9 (9 x)-1/2 (2 x)+c91logsec(9x)−21logsec(2x)+c, where c is the constant of integration

(9 x)- (2 x)+c, where c is the constant of integration

(9 x)+ (2 x)+c, where c is the constant of integration

1/9 (9 x)-1/2 (2 x)+c, where c is the constant of integration

1/9 (9 x)+1/2 (2 x)+c, where c is the constant of integration

Correct Answer is : 19log⁡sec⁡(9x)−12log⁡sec⁡(2x)+c1/9 (9 x)-1/2 (2 x)+c91logsec(9x)−21logsec(2x)+c, where c is the constant of integration

Test Index

MHTCET 5 May 2025 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int \frac{\sin 7 x}{\cos 9 x \cos 2 x} d x

is equal to

$\log \sec (9 x)-\log \sec (2 x)+c$ , where c is the constant of integration
$\log \sec (9 x)+\log \sec (2 x)+c$ , where c is the constant of integration
$\frac{1}{9} \log \sec (9 x)-\frac{1}{2} \log \sec (2 x)+c$ , where c is the constant of integration
$\frac{1}{9} \log \sec (9 x)+\frac{1}{2} \log \sec (2 x)+c$ , where c is the constant of integration

Solution:

Concept:

Use the identity

\sin(9x - 2x) = \sin 9x \cos 2x - \cos 9x \sin 2x

to split the integrand into

\tan 9x - \tan 2x

Explanation:

Since

7x = 9x - 2x

, rewrite the numerator:

\int \frac{\sin 7x}{\cos 9x \cos 2x}\,dx = \int \frac{\sin(9x - 2x)}{\cos 9x \cos 2x}\,dx

Using the sine difference formula:

= \int \frac{\sin 9x \cos 2x - \cos 9x \sin 2x}{\cos 9x \cos 2x}\,dx

Split into two fractions:

= \int \left( \frac{\sin 9x}{\cos 9x} - \frac{\sin 2x}{\cos 2x} \right)dx = \int (\tan 9x - \tan 2x)\,dx

Integrate term by term:

\int \tan 9x\,dx = \frac{1}{9}\log|\sec 9x| + c_1

\int \tan 2x\,dx = \frac{1}{2}\log|\sec 2x| + c_2

Therefore, the given integral equals:

\frac{1}{9}\log|\sec 9x| - \frac{1}{2}\log|\sec 2x| + c

Answer:

Option C:

\frac{1}{9} \log \sec (9x) - \frac{1}{2} \log \sec (2x) + c

, where

c

is the constant of integration.

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