NDA 2017
Exam NDA
Year 2017
Subject Mathematics
Chapter Definite Integration NDA 2017 · Mathematics · Definite Integration
The value of ∫_0^ (π)/(4) x d x+∫_0^ (π)/(4) x d x is equal to
The value of
∫ 0 π 4 tan x d x + ∫ 0 π 4 cot x d x \int_0^{\frac{\pi}{4}} \sqrt{\tan \mathrm{x}} d x+\int_0^{\frac{\pi}{4}} \sqrt{\cot \mathrm{x}} d x ∫ 0 4 π tan x d x + ∫ 0 4 π cot x d x is equal to
Options A π 4 \frac{\pi}{4} 4 π B π 2 \frac{\pi}{2} 2 π C π 2 2 \frac{\pi}{2 \sqrt{2}} 2 2 π D π 2 \frac{\pi}{\sqrt{2}} 2 π Correct answer D. π 2 \frac{\pi}{\sqrt{2}} 2 π
Step-by-step solution
∫ 0 π 4 tan x d x + ∫ 0 π 4 cot x ⋅ d x = ∫ 0 π 4 ( tan x + cot x ) d x \begin{aligned} & \int_0^{\frac{\pi}{4}} \sqrt{\tan x} d x+\int_0^{\frac{\pi}{4}} \sqrt{\cot x} \cdot d x \\ & =\int_0^{\frac{\pi}{4}}(\sqrt{\tan x}+\sqrt{\cot x}) d x\end{aligned} ∫ 0 4 π tan x d x + ∫ 0 4 π cot x ⋅ d x = ∫ 0 4 π ( tan x + cot x ) d x = ∫ 0 π 4 ( sin x cos x + cos x sin x ) d x = ∫ 0 π 4 sin x + cos x sin x ⋅ cos x ⋅ d x \begin{aligned} & =\int_0^{\frac{\pi}{4}}\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right) d x \\ & =\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{\sqrt{\sin x} \cdot \sqrt{\cos x}} \cdot d x\end{aligned} = ∫ 0 4 π ( cos x sin x + sin x cos x ) d x = ∫ 0 4 π sin x ⋅ cos x sin x + cos x ⋅ d x = 2 ∫ 0 π 4 ( sin x + cos x ) ( 1 − ( sin 2 x + cos 2 x − 2 sin x cos x ) ) ⋅ d x = 2 ∫ 0 π 4 ( sin x + cos x ) 1 − ( sin x − cos x ) 2 ⋅ d x \begin{aligned} & =\sqrt{2} \int_0^{\frac{\pi}{4}} \frac{(\sin x+\cos x)}{\left(\sqrt{1-\left(\sin ^2 x+\cos ^2 x-2 \sin x \cos x\right)}\right)} \cdot d x \\ & =\sqrt{2} \int_0^{\frac{\pi}{4}} \frac{(\sin x+\cos x)}{\sqrt{1-(\sin x-\cos x)^2}} \cdot d x\end{aligned} = 2 ∫ 0 4 π ( 1 − ( sin 2 x + cos 2 x − 2 sin x cos x ) ) ( sin x + cos x ) ⋅ d x = 2 ∫ 0 4 π 1 − ( sin x − cos x ) 2 ( sin x + cos x ) ⋅ d x
Put
sin x − cos \sin \mathrm{x}-\cos sin x − cos \mathrm{x}=\mathrm{t}
⇒ ( cos x + sin x ) d d = d t \Rightarrow(\cos \mathrm{x}+\sin \mathrm{x}) d \mathrm{ d}=\mathrm{dt} ⇒ ( cos x + sin x ) d d = dt When x=0, t=-1 and
x = π 4 , t = 0 x=\frac{\pi}{4}, t=0 x = 4 π , t = 0 .
= 2 ∫ − 1 0 1 1 − t 2 ⋅ d t = 2 ( sin − 1 t ) − 1 0 = 2 [ sin − 1 ( 0 ) − sin − 1 ( − 1 ) ] = 2 [ 0 − ( − π 2 ) ] = π 2 \begin{aligned} & =\sqrt{2} \int_{-1}^0 \frac{1}{\sqrt{1-\mathrm{t}^2}} \cdot \mathrm{dt}=\sqrt{2}\left(\sin ^{-1} \mathrm{t}\right)_{-1}^0 \\ & =\sqrt{2}\left[\sin ^{-1}(0)-\sin ^{-1}(-1)\right] \\ & =\sqrt{2}\left[0-\left(\frac{-\pi}{2}\right)\right]=\frac{\pi}{\sqrt{2}}\end{aligned} = 2 ∫ − 1 0 1 − t 2 1 ⋅ dt = 2 ( sin − 1 t ) − 1 0 = 2 [ sin − 1 ( 0 ) − sin − 1 ( − 1 ) ] = 2 [ 0 − ( 2 − π ) ] = 2 π
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