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Definite Integration question

2021 · 27 Jul · Shift 2 · Q39
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  5. /2021 · 27 Jul · Shift 2 · Q39

Definite Integration question

2021 · 27 Jul · Shift 2 · Q39

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
If ∫0π(sin⁡3x)e−sin⁡2xdx=α−βe∫01tetdt\int_0^\pi {({{\sin }^3}x){e^{ - {{\sin }^2}x}}dx = \alpha - {\beta \over e}\int_0^1 {\sqrt t {e^t}dt} }∫0π​(sin3x)e−sin2xdx=α−eβ​∫01​t​etdt, then α\alphaα+β\betaβ is equal to ‾\underline{\hspace{2cm}}​.
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Correct answer: 5

  1. We need to evaluate I=∫0πsin⁡3x e−sin⁡2x dx.I=\int_0^\pi \sin^3 x\, e^{-\sin^2 x}\,dx.I=∫0π​sin3xe−sin2xdx.

  2. Use sin⁡3x=sin⁡x(1−cos⁡2x).\sin^3 x=\sin x(1-\cos^2 x).sin3x=sinx(1−cos2x). So, I=∫0πsin⁡x(1−cos⁡2x)e−sin⁡2x dx.I=\int_0^\pi \sin x(1-\cos^2 x)e^{-\sin^2 x}\,dx.I=∫0π​sinx(1−cos2x)e−sin2xdx.

Also, sin⁡2x=1−cos⁡2x,\sin^2 x=1-\cos^2 x,sin2x=1−cos2x, therefore e−sin⁡2x=e−(1−cos⁡2x)=1eecos⁡2x.e^{-\sin^2 x}=e^{-(1-\cos^2 x)}=\frac{1}{e}e^{\cos^2 x}.e−sin2x=e−(1−cos2x)=e1​ecos2x.

Hence, I=1e∫0πsin⁡x(1−cos⁡2x)ecos⁡2x dx.I=\frac{1}{e}\int_0^\pi \sin x(1-\cos^2 x)e^{\cos^2 x}\,dx.I=e1​∫0π​sinx(1−cos2x)ecos2xdx.

  1. Put u=cos⁡x⇒du=−sin⁡x dx.u=\cos x \quad\Rightarrow\quad du=-\sin x\,dx.u=cosx⇒du=−sinxdx. When x=0x=0x=0, u=1u=1u=1. When x=πx=\pix=π, u=−1u=-1u=−1.

Thus,

=\frac{1}{e}\int_{-1}^{1} (1-u^2)e^{u^2}\,du.$$ 4. Since the integrand is even, $$I=\frac{2}{e}\int_0^1 (1-u^2)e^{u^2}\,du.$$ Now split: $$I=\frac{2}{e}\left(\int_0^1 e^{u^2}\,du-\int_0^1 u^2e^{u^2}\,du\right).$$ 5. Evaluate $$\int_0^1 u^2e^{u^2}\,du$$ using integration by parts via $$\frac{d}{du}(u e^{u^2})=e^{u^2}+2u^2e^{u^2}.$$ So, $$2u^2e^{u^2}=\frac{d}{du}(u e^{u^2})-e^{u^2},$$ which gives $$\int_0^1 u^2e^{u^2}\,du=\frac12\left[u e^{u^2}\right]_0^1-\frac12\int_0^1 e^{u^2}\,du =\frac{e}{2}-\frac12\int_0^1 e^{u^2}\,du.$$ Therefore, $$\int_0^1 e^{u^2}\,du-\int_0^1 u^2e^{u^2}\,du =\int_0^1 e^{u^2}\,du-\left(\frac e2-\frac12\int_0^1 e^{u^2}\,du\right) =\frac32\int_0^1 e^{u^2}\,du-\frac e2.$$ So, $$I=\frac{2}{e}\left(\frac32\int_0^1 e^{u^2}\,du-\frac e2\right) =\frac{3}{e}\int_0^1 e^{u^2}\,du-1.$$ 6. Now convert to the required form involving $\int_0^1 \sqrt t\,e^t\,dt$. Let $$t=u^2 \quad\Rightarrow\quad du=\frac{dt}{2\sqrt t}.$$ Then $$\int_0^1 e^{u^2}\,du=\frac12\int_0^1 t^{-1/2}e^t\,dt.$$ But a better relation is obtained directly for $$J=\int_0^1 \sqrt t\,e^t\,dt.$$ Integrate by parts: Take $$u=\sqrt t,\quad dv=e^t dt,$$ so $$du=\frac{1}{2\sqrt t}dt,\quad v=e^t.$$ Thus, $$J=\left[\sqrt t\,e^t\right]_0^1-\frac12\int_0^1 t^{-1/2}e^t\,dt =e-\int_0^1 e^{u^2}\,du,$$ since $$\int_0^1 t^{-1/2}e^t\,dt=2\int_0^1 e^{u^2}\,du.$$ Hence, $$\int_0^1 e^{u^2}\,du=e-J.$$ 7. Substitute into $I$: $$I=\frac{3}{e}(e-J)-1=3-\frac{3}{e}J-1=2-\frac{3}{e}\int_0^1 \sqrt t\,e^t\,dt.$$ So comparing with $$\int_0^\pi \sin^3x\,e^{-\sin^2x}dx=\alpha-\frac{\beta}{e}\int_0^1\sqrt t\,e^t\,dt,$$ we get $$\alpha=2,\quad \beta=3.$$ Therefore, $$\alpha+\beta=5.$$
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