Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Definite Integration question

2018 · 15 Apr · Shift 2 · Q40
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Definite Integration
  5. /2018 · 15 Apr · Shift 2 · Q40

Definite Integration question

2018 · 15 Apr · Shift 2 · Q40

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of integral ∫π43π4x1+sin⁡xdx\int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx}∫4π​43π​​1+sinxx​dx is :
  1. A
    π2\pi \sqrt 2π2​
  2. B
    π(2−1)\pi \left( {\sqrt 2 - 1} \right)π(2​−1)
  3. C
    π2(2+1){\pi \over 2}\left( {\sqrt 2 + 1} \right)2π​(2​+1)
  4. D
    2π(2−1)2\pi \left( {\sqrt 2 - 1} \right)2π(2​−1)
View written solutionFree

Correct answer: B

  1. Let I=∫π/43π/4x1+sin⁡x dx.I=\int_{\pi/4}^{3\pi/4}\frac{x}{1+\sin x}\,dx.I=∫π/43π/4​1+sinxx​dx.

We use the property ∫abf(x) dx=∫abf(a+b−x) dx.\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx.∫ab​f(x)dx=∫ab​f(a+b−x)dx. Here, a=π/4a=\pi/4a=π/4, b=3π/4b=3\pi/4b=3π/4, so a+b=πa+b=\pia+b=π.

Thus, I=∫π/43π/4π−x1+sin⁡(π−x) dx.I=\int_{\pi/4}^{3\pi/4}\frac{\pi-x}{1+\sin(\pi-x)}\,dx.I=∫π/43π/4​1+sin(π−x)π−x​dx. Since sin⁡(π−x)=sin⁡x\sin(\pi-x)=\sin xsin(π−x)=sinx, this becomes I=∫π/43π/4π−x1+sin⁡x dx.I=\int_{\pi/4}^{3\pi/4}\frac{\pi-x}{1+\sin x}\,dx.I=∫π/43π/4​1+sinxπ−x​dx.

  1. Add the two expressions for III:
=\pi\int_{\pi/4}^{3\pi/4}\frac{dx}{1+\sin x}.$$ So, $$I=\frac{\pi}{2}\int_{\pi/4}^{3\pi/4}\frac{dx}{1+\sin x}.$$ 3. Simplify the integrand: $$\frac{1}{1+\sin x}=\frac{1-\sin x}{1-\sin^2 x}=\frac{1-\sin x}{\cos^2 x} =\sec^2 x-\sec x\tan x.$$ Hence, $$\int \frac{dx}{1+\sin x}=\int (\sec^2 x-\sec x\tan x)\,dx =\tan x-\sec x.$$ 4. Evaluate from $\pi/4$ to $3\pi/4$: $$\int_{\pi/4}^{3\pi/4}\frac{dx}{1+\sin x} =\Big[\tan x-\sec x\Big]_{\pi/4}^{3\pi/4}.$$ Now, $$\tan\frac{3\pi}{4}=-1,\qquad \sec\frac{3\pi}{4}=-\sqrt2,$$ so $$\tan\frac{3\pi}{4}-\sec\frac{3\pi}{4}=-1-(-\sqrt2)=\sqrt2-1.$$ Also, $$\tan\frac{\pi}{4}=1,\qquad \sec\frac{\pi}{4}=\sqrt2,$$ so $$\tan\frac{\pi}{4}-\sec\frac{\pi}{4}=1-\sqrt2.$$ Therefore, $$\int_{\pi/4}^{3\pi/4}\frac{dx}{1+\sin x}=(\sqrt2-1)-(1-\sqrt2)=2(\sqrt2-1).$$ 5. Hence, $$I=\frac{\pi}{2}\cdot 2(\sqrt2-1)=\pi(\sqrt2-1).$$ So the correct option is $$\boxed{\text{B }\; \pi(\sqrt2-1)}.$$
PreviousNext

More from Definite Integration

  • If f(x)=0∫x​t(sinx−sint)dt then :2018 · MCQ
  • The value of −π/2∫π/2​1+2xsin2x​dx is2018 · MCQ
  • The integral ∫12π​4π​​(tanx+cotx)38cos2x​dx equals :2017 · MCQ
  • If 1∫2​(x2−2x+4)23​dx​=k+5k​, then k is equal to :2017 · MCQ
  • The integral 4π​∫43π​​1+cosxdx​ is equal to2017 · MCQ
  • If 20∫1​tan−1xdx=0∫1​cot−1(1−x+x2)dx, then 0∫1​tan−1(1−x+x2)dx is equalto :2016 · MCQ
  • For x ∈ R, x e 0, if y(x) is a differentiable function such that x 1∫x​y(t) dt = (x + 1) 1∫x​ty (t) dt, then y (x) equals : (where C is a constant.)2016 · MCQ
  • The value of the integral 4∫10​[x2−28x+196]+[x2][x2]dx​, where [x] denotes the greatest integer less than or equal to x, is :2016 · MCQ