JEE MAIN - Mathematics (2018 (Offline) - No. 13)
The integral
$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx$$
is equal to
$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx$$
is equal to
$${{ - 1} \over {1 + {{\cot }^3}x}} + C$$
$${1 \over {3\left( {1 + {{\tan }^3}x} \right)}} + C$$
$${{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + C$$
$${1 \over {1 + {{\cot }^3}x}} + C$$
Explanation
Given,
$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}}\,dx $$
$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left[ {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]}^2}}}} \,dx$$
$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}} dx$$
$$ = \int {{{{{\sin }^2}x\,\,{{\cos }^2}x} \over {{{\cos }^6}x{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,\,dx$$
$$ = \int {{{{{\sin }^2}x} \over {{{\cos }^4}x{{\left( {1 + {{\tan }^3}} \right)}^2}}}} \,\,dx$$
$$ = \int {{{{{\tan }^2}x.{{\sec }^2}x} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,dx$$
[ Let $$\,\,\,\,\,\,{\tan ^3}x = t$$
$$\,\,\,\,\,\,$$ $$ \Rightarrow 3{\tan ^2}x\,{\sec ^2}x\,\,dx = dt$$ ]
$$ = {1 \over 3}\int {{{3{{\tan }^2}x\,\,{{\sec }^2}x\,\,dx} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} $$
$$ = {1 \over 3}\int {{{dt} \over {{{\left( {1 + t} \right)}^2}}}} $$
$$ = {1 \over 3} \times {{{{\left( {1 + t} \right)}^{ - 2 + 1}}} \over { - 2 + 1}} + c$$
$$ = {1 \over 3} \times {{ - 1} \over {\left( {1 + t} \right)}} + c$$
$$ = {{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + c$$
$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}}\,dx $$
$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left[ {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]}^2}}}} \,dx$$
$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}} dx$$
$$ = \int {{{{{\sin }^2}x\,\,{{\cos }^2}x} \over {{{\cos }^6}x{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,\,dx$$
$$ = \int {{{{{\sin }^2}x} \over {{{\cos }^4}x{{\left( {1 + {{\tan }^3}} \right)}^2}}}} \,\,dx$$
$$ = \int {{{{{\tan }^2}x.{{\sec }^2}x} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,dx$$
[ Let $$\,\,\,\,\,\,{\tan ^3}x = t$$
$$\,\,\,\,\,\,$$ $$ \Rightarrow 3{\tan ^2}x\,{\sec ^2}x\,\,dx = dt$$ ]
$$ = {1 \over 3}\int {{{3{{\tan }^2}x\,\,{{\sec }^2}x\,\,dx} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} $$
$$ = {1 \over 3}\int {{{dt} \over {{{\left( {1 + t} \right)}^2}}}} $$
$$ = {1 \over 3} \times {{{{\left( {1 + t} \right)}^{ - 2 + 1}}} \over { - 2 + 1}} + c$$
$$ = {1 \over 3} \times {{ - 1} \over {\left( {1 + t} \right)}} + c$$
$$ = {{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + c$$
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