JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 17)
If $$\int_0^\pi {({{\sin }^3}x){e^{ - {{\sin }^2}x}}dx = \alpha - {\beta \over e}\int_0^1 {\sqrt t {e^t}dt} } $$, then $$\alpha$$ + $$\beta$$ is equal to ____________.
Answer
5
Explanation
$$I = 2\int_0^{\pi /2} {{{\sin }^3}x{e^{ - {{\sin }^2}x}}dx} $$
$$ = 2\int_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \int\limits_0^{\pi /2} {\mathop {\cos x}\limits_I \,\underbrace {{e^{ - {{\sin }^2}x}}( - \sin 2x)}_{II}dx} $$$$ = 2\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \left[ {\cos x{e^{ - {{\sin }^2}x}}} \right]_0^{\pi /2} + \int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}} dx$$
$$ = 3\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} - 1$$
$$ = {3 \over 2}\int\limits_{ - 1}^0 {{{{e^\alpha }d\alpha } \over {\sqrt {1 + \alpha } }} - 1} $$ (Put $$-$$sin2x = t)
$$ = {3 \over {2e}}\int\limits_0^1 {{{{e^x}} \over {\sqrt x }}dx - 1} $$ (put 1 + $$\alpha$$ = x)
$$ = {3 \over {2e}}\int\limits_0^1 {\mathop {{e^x}}\limits_{II} } \mathop {{1 \over {\sqrt x }}}\limits_{II} dx - 1$$
$$ = 2 - {3 \over e}\int\limits_0^1 {{e^x}\sqrt x } dx$$
Hence, $$\alpha$$ + $$\beta$$ = 5
$$ = 2\int_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \int\limits_0^{\pi /2} {\mathop {\cos x}\limits_I \,\underbrace {{e^{ - {{\sin }^2}x}}( - \sin 2x)}_{II}dx} $$$$ = 2\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \left[ {\cos x{e^{ - {{\sin }^2}x}}} \right]_0^{\pi /2} + \int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}} dx$$
$$ = 3\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} - 1$$
$$ = {3 \over 2}\int\limits_{ - 1}^0 {{{{e^\alpha }d\alpha } \over {\sqrt {1 + \alpha } }} - 1} $$ (Put $$-$$sin2x = t)
$$ = {3 \over {2e}}\int\limits_0^1 {{{{e^x}} \over {\sqrt x }}dx - 1} $$ (put 1 + $$\alpha$$ = x)
$$ = {3 \over {2e}}\int\limits_0^1 {\mathop {{e^x}}\limits_{II} } \mathop {{1 \over {\sqrt x }}}\limits_{II} dx - 1$$
$$ = 2 - {3 \over e}\int\limits_0^1 {{e^x}\sqrt x } dx$$
Hence, $$\alpha$$ + $$\beta$$ = 5
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