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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 129 of 150

Marks: +1, -0

\int_{\pi / 6}^{\pi / 3} \frac{\sin x-\cos x}{1+\sin x \cos x} d x=

Solution:

Concept:

Use the substitution

t = \sin x + \cos x

, because its derivative is directly related to the numerator.

Explanation:

Let

t = \sin x + \cos x

Then

dt = (\cos x - \sin x)\,dx = -(\sin x - \cos x)\,dx

Now,

t^2 = \sin^2 x + \cos^2 x + 2\sin x\cos x = 1 + 2\sin x\cos x

So,

1 + \sin x\cos x = 1 + \frac{t^2 - 1}{2} = \frac{t^2 + 1}{2}

Substituting into the integral gives:

\int \frac{\sin x - \cos x}{1+\sin x\cos x}\,dx = \int \frac{-dt}{\frac{t^2 + 1}{2}} = -2\tan^{-1}(t).

Thus, the antiderivative is

-2\tan^{-1}(\sin x + \cos x)

Evaluate at the limits:

\sin\frac{\pi}{6} + \cos\frac{\pi}{6} = \frac{1+\sqrt{3}}{2}, \quad \sin\frac{\pi}{3} + \cos\frac{\pi}{3} = \frac{\sqrt{3}+1}{2}.

Both endpoint values are equal, so the definite integral becomes

0

Answer:

Option A:

0

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