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Indefinite Integrals question

2025 · 23 Jan · Shift 2 · Q42
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Indefinite Integrals question

2025 · 23 Jan · Shift 2 · Q42

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
Let ∫x3sin⁡x dx=g(x)+C\int x^3 \sin x \mathrm{~d} x=g(x)+C∫x3sinx dx=g(x)+C, where CCC is the constant of integration. If 8(g(π2)+g′(π2))=απ3+βπ2+γ,α,β,γ∈Z8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z8(g(2π​)+g′(2π​))=απ3+βπ2+γ,α,β,γ∈Z, then α+β−γ\alpha+\beta-\gammaα+β−γ equals :
  1. A
    47
  2. B
    55
  3. C
    62
  4. D
    48
View written solutionFree

Correct answer: B

  1. We are given ∫x3sin⁡x dx=g(x)+C\int x^3\sin x\,dx=g(x)+C∫x3sinxdx=g(x)+C So, by definition, g′(x)=x3sin⁡x.g'(x)=x^3\sin x.g′(x)=x3sinx.

We need to compute 8(g(π2)+g′(π2)).8\left(g\left(\frac{\pi}{2}\right)+g'\left(\frac{\pi}{2}\right)\right).8(g(2π​)+g′(2π​)).

  1. First find one antiderivative of x3sin⁡xx^3\sin xx3sinx.

Use integration by parts repeatedly: I=∫x3sin⁡x dxI=\int x^3\sin x\,dxI=∫x3sinxdx Take u=x3,dv=sin⁡x dxu=x^3,\quad dv=\sin x\,dxu=x3,dv=sinxdx Then du=3x2dx,v=−cos⁡xdu=3x^2dx,\quad v=-\cos xdu=3x2dx,v=−cosx Hence, I=−x3cos⁡x+3∫x2cos⁡x dx.I=-x^3\cos x+3\int x^2\cos x\,dx.I=−x3cosx+3∫x2cosxdx.

Now let J=∫x2cos⁡x dxJ=\int x^2\cos x\,dxJ=∫x2cosxdx Again by parts: u=x2,dv=cos⁡x dxu=x^2,\quad dv=\cos x\,dxu=x2,dv=cosxdx So du=2x dx,v=sin⁡xdu=2x\,dx,\quad v=\sin xdu=2xdx,v=sinx Thus, J=x2sin⁡x−2∫xsin⁡x dx.J=x^2\sin x-2\int x\sin x\,dx.J=x2sinx−2∫xsinxdx.

Now let K=∫xsin⁡x dxK=\int x\sin x\,dxK=∫xsinxdx Again by parts: u=x,dv=sin⁡x dxu=x,\quad dv=\sin x\,dxu=x,dv=sinxdx So du=dx,v=−cos⁡xdu=dx,\quad v=-\cos xdu=dx,v=−cosx Hence, K=−xcos⁡x+∫cos⁡x dx=−xcos⁡x+sin⁡x.K=-x\cos x+\int \cos x\,dx=-x\cos x+\sin x.K=−xcosx+∫cosxdx=−xcosx+sinx.

Therefore, J=x2sin⁡x−2(−xcos⁡x+sin⁡x)J=x^2\sin x-2(-x\cos x+\sin x)J=x2sinx−2(−xcosx+sinx) J=x2sin⁡x+2xcos⁡x−2sin⁡x.J=x^2\sin x+2x\cos x-2\sin x.J=x2sinx+2xcosx−2sinx.

Substitute into III: I=−x3cos⁡x+3(x2sin⁡x+2xcos⁡x−2sin⁡x)I=-x^3\cos x+3\left(x^2\sin x+2x\cos x-2\sin x\right)I=−x3cosx+3(x2sinx+2xcosx−2sinx) I=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x.I=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x.I=−x3cosx+3x2sinx+6xcosx−6sinx.

So we can take g(x)=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6sin⁡x.g(x)=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x.g(x)=−x3cosx+3x2sinx+6xcosx−6sinx.

  1. Now evaluate g(π2)g\left(\frac{\pi}{2}\right)g(2π​). Using sin⁡π2=1,cos⁡π2=0,\sin\frac{\pi}{2}=1,\qquad \cos\frac{\pi}{2}=0,sin2π​=1,cos2π​=0, we get g(π2)=−(π2)3(0)+3(π2)2(1)+6(π2)(0)−6(1).g\left(\frac{\pi}{2}\right)= -\left(\frac{\pi}{2}\right)^3(0)+3\left(\frac{\pi}{2}\right)^2(1)+6\left(\frac{\pi}{2}\right)(0)-6(1).g(2π​)=−(2π​)3(0)+3(2π​)2(1)+6(2π​)(0)−6(1). Thus, g(π2)=3⋅π24−6=3π24−6.g\left(\frac{\pi}{2}\right)=3\cdot\frac{\pi^2}{4}-6=\frac{3\pi^2}{4}-6.g(2π​)=3⋅4π2​−6=43π2​−6.

  2. Now evaluate g′(π2)g'\left(\frac{\pi}{2}\right)g′(2π​). Since g′(x)=x3sin⁡x,g'(x)=x^3\sin x,g′(x)=x3sinx, we get g′(π2)=(π2)3⋅1=π38.g'\left(\frac{\pi}{2}\right)=\left(\frac{\pi}{2}\right)^3\cdot 1=\frac{\pi^3}{8}.g′(2π​)=(2π​)3⋅1=8π3​.

  3. Therefore, g(π2)+g′(π2)=3π24−6+π38.g\left(\frac{\pi}{2}\right)+g'\left(\frac{\pi}{2}\right)=\frac{3\pi^2}{4}-6+\frac{\pi^3}{8}.g(2π​)+g′(2π​)=43π2​−6+8π3​. Multiply by 888: 8(g(π2)+g′(π2))=8(3π24−6+π38).8\left(g\left(\frac{\pi}{2}\right)+g'\left(\frac{\pi}{2}\right)\right)=8\left(\frac{3\pi^2}{4}-6+\frac{\pi^3}{8}\right).8(g(2π​)+g′(2π​))=8(43π2​−6+8π3​). So, =6π2−48+π3.=6\pi^2-48+\pi^3.=6π2−48+π3.

Comparing with απ3+βπ2+γ,\alpha\pi^3+\beta\pi^2+\gamma,απ3+βπ2+γ, we get α=1,β=6,γ=−48.\alpha=1,\quad \beta=6,\quad \gamma=-48.α=1,β=6,γ=−48.

  1. Now compute α+β−γ=1+6−(−48)=55.\alpha+\beta-\gamma=1+6-(-48)=55.α+β−γ=1+6−(−48)=55.

  2. Hence the correct option is B: 55.\boxed{\text{B: }55}.B: 55​.

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