JEE Main 2021 — Mathematics Calculus

Let, $\displaystyle I={\int }_{0}^{100\pi }\frac{{\mathrm{sin}}^{2}x}{{e}^{(\frac{x}{\pi }-[\frac{x}{\pi }])}}dx$I=∫0100π​e(πx​−[πx​])sin2x​dx

We know that if $f(x)$f(x) is a periodic function, with period $T$T then $\displaystyle {\int }_{0}^{nT}f(x)dx=n{\int }_{0}^{T}f(x)dx,n\in Z$∫0nT​f(x)dx=n∫0T​f(x)dx,n∈Z

$$\Rightarrow I=100{\int }_{0}^{\pi }\frac{{\mathrm{sin}}^{2}x}{{e}^{(\frac{x}{\pi }-[\frac{x}{\pi }])}}dx$$⇒I=100∫0π​e(πx​−[πx​])sin2x​dx

For, $0<x<\pi ,0<\frac{x}{\pi }<1$0<x<π,0<πx​<1 $\Rightarrow [\frac{x}{\pi }]=0$⇒[πx​]=0

$$\Rightarrow I=100{\int }_{0}^{\pi }\frac{{\mathrm{sin}}^{2}x}{{e}^{(\frac{x}{\pi })}}dx$$⇒I=100∫0π​e(πx​)sin2x​dx

Now, using $\mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x,$cos2x=1−2sin2x, we get

$$I=100{\int }_{0}^{\pi }{e}^{-x/\pi }\frac{(1-\mathrm{cos}2x)}{2}dx$$I=100∫0π​e−x/π2(1−cos2x)​dx

$$\Rightarrow I=100\times \frac{1}{2}{\int }_{0}^{\pi }({e}^{-x/\pi }-{e}^{-x/\pi }\cdot \mathrm{cos}2x)dx$$⇒I=100×21​∫0π​(e−x/π−e−x/π⋅cos2x)dx

$$\Rightarrow I=50{{\int }_{0}^{\pi }{e}^{-x/\pi }dx-{\int }_{0}^{\pi }{e}^{-x/\pi }\cdot \mathrm{cos}2xdx}$$⇒I=50∫0π​e−x/πdx−∫0π​e−x/π⋅cos2xdx

Let, $\displaystyle {I}_{1}={\int }_{0}^{\pi }{e}^{-x/\pi }dx$I1​=∫0π​e−x/πdx

$$\Rightarrow {I}_{1}={[-\pi {e}^{-x/\pi }]}_{0}^{\pi }$$⇒I1​=[−πe−x/π]0π​

$$\Rightarrow {I}_{1}=-\pi ({e}^{-\pi /\pi }-{e}^{0})=\pi (1-{e}^{-1})$$⇒I1​=−π(e−π/π−e0)=π(1−e−1)

And, let $\displaystyle {I}_{2}={\int }_{0}^{\pi }{e}^{-x/\pi }\cdot \mathrm{cos}2xdx$I2​=∫0π​e−x/π⋅cos2xdx

By using $ILATE$ILATE rule,

$${I}_{2}={\int }_{0}^{\pi }\mathrm{cos}2x\cdot {e}^{-x/\pi }dx$$I2​=∫0π​cos2x⋅e−x/πdx

Now, using integration by parts, i.e. $\displaystyle \int (u\cdot v)dx=u\int vdx-\int [\frac{d}{dx}(u)\int vdx]dx,$∫(u⋅v)dx=u∫vdx−∫[dxd​(u)∫vdx]dx, we get

$$\Rightarrow {I}_{2}=-\pi {[{e}^{-x/\pi }\cdot \mathrm{cos}2x]}_{0}^{\pi }-{\int }_{0}^{\pi }(-\pi {e}^{-x/\pi }\cdot (-2\mathrm{sin}2x))dx$$⇒I2​=−π[e−x/π⋅cos2x]0π​−∫0π​(−πe−x/π⋅(−2sin2x))dx

$$\Rightarrow {I}_{2}=\pi (1-{e}^{-1})-2\pi {\int }_{0}^{\pi }{e}^{-x/\pi }\mathrm{sin}2xdx$$⇒I2​=π(1−e−1)−2π∫0π​e−x/πsin2xdx

Again, applying integration by parts, we get

$${I}_{2}=\pi (1-{e}^{-1})-2\pi {{[-\pi {e}^{-x/\pi }\mathrm{sin}2x]}_{0}^{\pi }-{\int }_{0}^{\pi }-\pi {e}^{-x/\pi }2\mathrm{cos}2xdx}$$I2​=π(1−e−1)−2π[−πe−x/πsin2x]0π​−∫0π​−πe−x/π2cos2xdx

$$\Rightarrow {I}_{2}=\pi (1-{e}^{-1})-4{\pi }^{2}{I}_{2}$$⇒I2​=π(1−e−1)−4π2I2​

$$\Rightarrow {I}_{2}=\frac{\pi (1-{e}^{-1})}{1+4{\pi }^{2}}$$⇒I2​=1+4π2π(1−e−1)​

$$\therefore I=50({I}_{1}+{I}_{2})$$∴I=50(I1​+I2​)

$$\Rightarrow I=50{\pi (1-{e}^{-1})-\frac{\pi (1-{e}^{-1})}{1+4{\pi }^{2}}}$$⇒I=50π(1−e−1)−1+4π2π(1−e−1)​

$$\Rightarrow I=\frac{200(1-{e}^{-1}){\pi }^{3}}{1+4{\pi }^{2}}$$⇒I=1+4π2200(1−e−1)π3​

Given $I=\frac{\alpha {\pi }^{3}}{1+4{\pi }^{2}}$I=1+4π2απ3​

$$\Rightarrow \frac{\alpha {\pi }^{3}}{1+4{\pi }^{2}}=\frac{200(1-{e}^{-1}){\pi }^{3}}{1+4{\pi }^{2}}$$⇒1+4π2απ3​=1+4π2200(1−e−1)π3​

$$\Rightarrow \alpha =200(1-{e}^{-1}).$$⇒α=200(1−e−1).