QJEEM19Ma4386: JEE Main 2019 Mathematics - Class 12 JEEM19Ma4386 Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx}

is

Accepted Answer

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}

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