JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 5)
The value of $$\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} $$ is
$${{\pi - 2} \over 8}$$
$${{\pi - 2} \over 4}$$
$${{\pi - 1} \over 2}$$
$${{\pi - 1} \over 4}$$
Explanation
I = $$\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} $$ .....(1)
$$ \Rightarrow $$ I = $$\int\limits_0^{{\pi \over 2}} {{{{{\cos }^3}x} \over {\sin x + \cos x}}} dx$$ ......(2)
Adding those two
2I = $$\int\limits_0^{{\pi \over 2}} {{{si{n^3}x + {{\cos }^3}x} \over {\sin x + \cos x}}} dx$$
= $$\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x + \cos x} \right)\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} \over {\sin x + \cos x}}} dx$$
= $$\int\limits_0^{{\pi \over 2}} {\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} dx$$
= $$\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\sin 2x} \over 2}} \right)} dx$$
= $$\left[ {x + {{\cos 2x} \over 4}} \right]_0^{{\pi \over 2}}$$
= $$\left( {{\pi \over 2} - {1 \over 4}} \right) - \left( {{1 \over 4}} \right)$$
$$ \therefore $$ 2I = $$\left( {{\pi \over 2} - {1 \over 2}} \right)$$
$$ \therefore $$ I = $${{\pi - 1} \over 4}$$
$$ \Rightarrow $$ I = $$\int\limits_0^{{\pi \over 2}} {{{{{\cos }^3}x} \over {\sin x + \cos x}}} dx$$ ......(2)
Adding those two
2I = $$\int\limits_0^{{\pi \over 2}} {{{si{n^3}x + {{\cos }^3}x} \over {\sin x + \cos x}}} dx$$
= $$\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x + \cos x} \right)\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} \over {\sin x + \cos x}}} dx$$
= $$\int\limits_0^{{\pi \over 2}} {\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} dx$$
= $$\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\sin 2x} \over 2}} \right)} dx$$
= $$\left[ {x + {{\cos 2x} \over 4}} \right]_0^{{\pi \over 2}}$$
= $$\left( {{\pi \over 2} - {1 \over 4}} \right) - \left( {{1 \over 4}} \right)$$
$$ \therefore $$ 2I = $$\left( {{\pi \over 2} - {1 \over 2}} \right)$$
$$ \therefore $$ I = $${{\pi - 1} \over 4}$$
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