The value of { x · x}{9-16 ² x} {dx} is equal to

Correct Answer is : 140log⁡(5+4sec⁡x5−4sec⁡x)+c1/40 ≤ft(5+4 x/5-4 x)+c401log(5−4secx5+4secx)+c, (where c is a constant of integration)

1/24 ≤ft(5+4 x/5-4 x)+c, (where c is a constant of integration)

1/40 ≤ft(5+4 x/5-4 x)+c, (where c is a constant of integration)

1/24 ≤ft(5-4 x/5+4 x)+c, (where c is a constant of integration)

1/40 ≤ft(5-4 x/5+4 x)+c₁ (where c is a constant of integration)

Correct Answer is : 140log⁡(5+4sec⁡x5−4sec⁡x)+c1/40 ≤ft(5+4 x/5-4 x)+c401log(5−4secx5+4secx)+c, (where c is a constant of integration)

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MHTCET 15 May 2024 Shift 1 Paper

Question : 112 of 150

Marks: +1, -0

The value of

\int \frac{\sec x \cdot \tan x}{9-16 \tan ^{2} x} \mathrm{dx}

is equal to

$\frac{1}{24} \log \left(\frac{5+4 \sec x}{5-4 \sec x}\right)+c$ , (where c is a constant of integration)
$\frac{1}{40} \log \left(\frac{5+4 \sec x}{5-4 \sec x}\right)+c$ , (where c is a constant of integration)
$\frac{1}{24} \log \left(\frac{5-4 \sec x}{5+4 \sec x}\right)+c$ , (where c is a constant of integration)
$\frac{1}{40} \log \left(\frac{5-4 \sec x}{5+4 \sec x}\right)+c_{1}$ (where c is a constant of integration)

Solution:

Concept:

Use the substitution

t = \sec x

and the standard integral

\int \frac{dt}{a^2 - (bt)^2} = \frac{1}{2ab} \log \left|\frac{a+bt}{a-bt}\right| + c

Explanation:

Let

I = \int \frac{\sec x \tan x}{9 - 16\tan^2 x}\, dx

Use the identity

\tan^2 x = \sec^2 x - 1

Then

9 - 16\tan^2 x = 9 - 16(\sec^2 x - 1) = 25 - 16\sec^2 x = 5^2 - (4\sec x)^2

Put

t = \sec x

, so

dt = \sec x \tan x\, dx

Therefore

I = \int \frac{dt}{5^2 - (4t)^2}

Using the standard formula,

I = \frac{1}{2(5)(4)} \log \left|\frac{5+4t}{5-4t}\right| + c

Substitute back

t = \sec x

I = \frac{1}{40} \log \left|\frac{5+4\sec x}{5-4\sec x}\right| + c

This is equal to option B, where the modulus sign is omitted in the given option.

Answer:

Option B:

\frac{1}{40} \log \left(\frac{5+4\sec x}{5-4\sec x}\right)+c

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