₀^{π/2} { x dx}{{1+ x x}} =

Correct Answer is : 2sin⁡−1(13){2} ^{-1}≤ft({1}{{3}})2sin−1(31)

Correct Answer is : 2sin⁡−1(13){2} ^{-1}≤ft({1}{{3}})2sin−1(31)

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AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Section: Mathematics

Question : 77 of 160

Marks: +1, -0

\int\limits_{0}^{\pi/2}

\frac{\cos x dx}{\sqrt{1+\cos x \sin x}}

Solution:

Let,

I=\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos x dx}{\sqrt{1+\cos x \sin x}}

Use property

\int\limits_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a-x) dx

I=\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right) \sin\left(\frac{\pi}{2}-x\right)} dx

=\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos x}{1+\sin x \cos x} dx

Then,

2 I =\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos x+\sin x}{\sqrt{1+\cos x \sin x}} \times \frac{\sqrt{2}}{\sqrt{2}} dx

=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x+\sin x)}{\sqrt{2+2\cos x \sin x}} dx

=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sqrt{2}(\cos x+\sin x)}{\sqrt{3-(\sin x-\cos x)^{2}}} dx

Put

\sin x-\cos x = t

\Rightarrow (\cos x+\sin x) dx = dt

2 I =\int\limits_{-1}^{1} \frac{\sqrt{2}}{\sqrt{3-t^{2}}} dt

=2\int\limits_{0}^{1} \frac{\sqrt{2}}{\sqrt{3-t^{2}}} dt

I=\sqrt{2} \left[ \sin^{-1}\left(\frac{t}{\sqrt{3}}\right) \right]_{-1}^{1}

=\sqrt{2} \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)

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