Verified Solution • 2025–26 Session moderate

QJEEM21Mab461: JEE Main 2021 Mathematics - Class 12 JEEM21Mab461 Solution: If , then + is equal to ____________.

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} } , then α\alpha + β\beta is equal to ____________.

Step-by-Step Verified Solution

I=2∫0π/2sin⁡3xe−sin⁡2xdxI = 2\int_0^{\pi /2} {{{\sin }^3}x{e^{ - {{\sin }^2}x}}dx}

=2∫0π/2sin⁡xe−sin⁡2xdx+∫0π/2cos⁡xI e−sin⁡2x(−sin⁡2x)⏟IIdx= 2\int_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \int\limits_0^{\pi /2} {\mathop {\cos x}\limits_I \,\underbrace {{e^{ - {{\sin }^2}x}}( - \sin 2x)}_{II}dx} =2∫0π/2sin⁡xe−sin⁡2xdx+[cos⁡xe−sin⁡2x]0π/2+∫0π/2sin⁡xe−sin⁡2xdx= 2\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \left[ {\cos x{e^{ - {{\sin }^2}x}}} \right]_0^{\pi /2} + \int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}} dx

=3∫0π/2sin⁡xe−sin⁡2xdx−1= 3\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} - 1

=32∫−10eαdα1+α−1= {3 \over 2}\int\limits_{ - 1}^0 {{{{e^\alpha }d\alpha } \over {\sqrt {1 + \alpha } }} - 1} (Put −- sin2x = t)

=32e∫01exxdx−1= {3 \over {2e}}\int\limits_0^1 {{{{e^x}} \over {\sqrt x }}dx - 1} (put 1 + α\alpha = x)

=32e∫01exII1xIIdx−1= {3 \over {2e}}\int\limits_0^1 {\mathop {{e^x}}\limits_{II} } \mathop {{1 \over {\sqrt x }}}\limits_{II} dx - 1

=2−3e∫01exxdx= 2 - {3 \over e}\int\limits_0^1 {{e^x}\sqrt x } dx

Hence, α\alpha + β\beta = 5

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