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Mathematics · Definite Integration

JEE Main 2024 — 29 January, Shift 2 — Question 24

If ∫π6π31−sin⁡2xdx=α+β2+γ3\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}, where α\alpha, β\beta and γ\gamma are rational numbers, then 3α+4β−γ3 \alpha+4 \beta-\gamma is equal to _______\_\_\_\_\_\_\_ .

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

=∫π6π31−sin⁡2xdx=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x

=∫π6π3∣sin⁡x−cos⁡x∣dx=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}|\sin x-\cos x| d x

=∫π6π4(cos⁡x−sin⁡x)dx+∫π4π3(sin⁡x−cos⁡x)dx=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}(\sin x-\cos x) d x

=−1+22−3=-1+2 \sqrt{2}-\sqrt{3}

=α+β2+γ3=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}

α=−1,β=2,γ=−1\alpha=-1, \beta=2, \gamma=-1

3α+4β−γ=63 \alpha+4 \beta-\gamma=6

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Mathematics
Chapter
Definite Integration
Topic
Evaluation of Definite Integrals

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