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Indefinite Integrals question

2020 · 8 Jan · Shift 1 · Q27
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Indefinite Integrals question

2020 · 8 Jan · Shift 1 · Q27

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
If ∫cos⁡xdxsin⁡3x(1+sin⁡6x)2/3=f(x)(1+sin⁡6x)1/λ+c\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}} = f\left( x \right){\left( {1 + {{\sin }^6}x} \right)^{1/\lambda }} + c∫sin3x(1+sin6x)2/3cosxdx​=f(x)(1+sin6x)1/λ+c where c is a constant of integration, then λf(π3)\lambda f\left( {{\pi \over 3}} \right)λf(3π​) is equal to
  1. A
    98{9 \over 8}89​
  2. B
    2
  3. C
    -2
  4. D
    −98-{9 \over 8}−89​
View written solutionFree

Correct answer: C

  1. We need to evaluate
I=∫cos⁡x dxsin⁡3x (1+sin⁡6x)2/3I= \int \frac{\cos x\,dx}{\sin^3 x\,(1+\sin^6 x)^{2/3}}I=∫sin3x(1+sin6x)2/3cosxdx​

and match it with

I=f(x)(1+sin⁡6x)1/λ+c.I=f(x)(1+\sin^6 x)^{1/\lambda}+c.I=f(x)(1+sin6x)1/λ+c.
  1. Substitute
t=sin⁡x⇒dt=cos⁡x dx.t=\sin x \quad\Rightarrow\quad dt=\cos x\,dx.t=sinx⇒dt=cosxdx.

Then

I=∫dtt3(1+t6)2/3.I=\int \frac{dt}{t^3(1+t^6)^{2/3}}.I=∫t3(1+t6)2/3dt​.
  1. Now use the substitution
u=t−2.u=t^{-2}.u=t−2.

Then

dνdt=−2t−3⇒t−3dt=−12dν.\frac{d\nu}{dt}=-2t^{-3} \quad\Rightarrow\quad t^{-3}dt=-\frac12 d\nu.dtdν​=−2t−3⇒t−3dt=−21​dν.

Also,

1+t6=1+1ν3=ν3+1ν3.1+t^6=1+\frac{1}{\nu^3}=\frac{\nu^3+1}{\nu^3}.1+t6=1+ν31​=ν3ν3+1​.

Hence

(1+t6)2/3=(ν3+1)2/3ν2.(1+t^6)^{2/3}=\frac{(\nu^3+1)^{2/3}}{\nu^2}.(1+t6)2/3=ν2(ν3+1)2/3​.

So

I=∫t−3dt(1+t6)2/3=∫(−12dν)ν2(ν3+1)2/3.I=\int \frac{t^{-3}dt}{(1+t^6)^{2/3}} =\int \left(-\frac12 d\nu\right)\frac{\nu^2}{(\nu^3+1)^{2/3}}.I=∫(1+t6)2/3t−3dt​=∫(−21​dν)(ν3+1)2/3ν2​.

Thus

I=−12∫ν2 dν(1+ν3)2/3.I=-\frac12\int \frac{\nu^2\,d\nu}{(1+\nu^3)^{2/3}}.I=−21​∫(1+ν3)2/3ν2dν​.
  1. Let
w=1+ν3.w=1+\nu^3.w=1+ν3.

Then

dw=3ν2dν⇒ν2dν=dw3.dw=3\nu^2 d\nu \quad\Rightarrow\quad \nu^2 d\nu=\frac{dw}{3}.dw=3ν2dν⇒ν2dν=3dw​.

Therefore

I=−12⋅13∫w−2/3dw=−16⋅w1/31/3+c=−12w1/3+c.I=-\frac12\cdot \frac13 \int w^{-2/3}dw =-\frac16\cdot \frac{w^{1/3}}{1/3}+c =-\frac12 w^{1/3}+c.I=−21​⋅31​∫w−2/3dw=−61​⋅1/3w1/3​+c=−21​w1/3+c.

Back-substituting,

I=−12(1+ν3)1/3+c.I=-\frac12(1+\nu^3)^{1/3}+c.I=−21​(1+ν3)1/3+c.

Since

ν=t−2=sin⁡−2x,\nu=t^{-2}=\sin^{-2}x,ν=t−2=sin−2x,

we get

I=−12(1+sin⁡−6x)1/3+c.I=-\frac12\left(1+\sin^{-6}x\right)^{1/3}+c.I=−21​(1+sin−6x)1/3+c.
  1. Rewrite in the required form. Note that
1+sin⁡−6x=1+sin⁡6xsin⁡6x.1+\sin^{-6}x=\frac{1+\sin^6 x}{\sin^6 x}.1+sin−6x=sin6x1+sin6x​.

Hence

(1+sin⁡−6x)1/3=(1+sin⁡6x)1/3sin⁡2x.\left(1+\sin^{-6}x\right)^{1/3} =\frac{(1+\sin^6 x)^{1/3}}{\sin^2 x}.(1+sin−6x)1/3=sin2x(1+sin6x)1/3​.

Therefore

I=−12sin⁡2x(1+sin⁡6x)1/3+c.I=-\frac{1}{2\sin^2 x}(1+\sin^6 x)^{1/3}+c.I=−2sin2x1​(1+sin6x)1/3+c.

Comparing with

I=f(x)(1+sin⁡6x)1/λ+c,I=f(x)(1+\sin^6 x)^{1/\lambda}+c,I=f(x)(1+sin6x)1/λ+c,

we get

f(x)=−12sin⁡2x,1λ=13.f(x)=-\frac{1}{2\sin^2 x}, \qquad \frac{1}{\lambda}=\frac13.f(x)=−2sin2x1​,λ1​=31​.

So

λ=3.\lambda=3.λ=3.
  1. Now compute f(π3)f\left(\frac\pi3\right)f(3π​):
sin⁡π3=32⇒sin⁡2π3=34.\sin\frac\pi3=\frac{\sqrt3}{2} \quad\Rightarrow\quad \sin^2\frac\pi3=\frac34.sin3π​=23​​⇒sin23π​=43​.

Thus

f(π3)=−12⋅34=−13/2=−23.f\left(\frac\pi3\right)=-\frac{1}{2\cdot \frac34}=-\frac{1}{3/2}=-\frac23.f(3π​)=−2⋅43​1​=−3/21​=−32​.

Therefore

λf(π3)=3(−23)=−2.\lambda f\left(\frac\pi3\right)=3\left(-\frac23\right)=-2.λf(3π​)=3(−32​)=−2.
  1. Hence the correct option is
C (−2).\boxed{\text{C }(-2)}.C (−2)​.
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