JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 8)
If $$\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}} = f\left( x \right){\left( {1 + {{\sin }^6}x} \right)^{1/\lambda }} + c$$
where c is a constant of integration, then $$\lambda f\left( {{\pi \over 3}} \right)$$ is equal to
where c is a constant of integration, then $$\lambda f\left( {{\pi \over 3}} \right)$$ is equal to
$${9 \over 8}$$
2
-2
$$-{9 \over 8}$$
Explanation
Given I = $$\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}}$$
Let sin x = t
$$ \Rightarrow $$ cos xdx = dt
$$ \therefore $$ I = $$\int {{{dt} \over {{t^3}{{\left( {1 + {t^6}} \right)}^{2/3}}}}} $$
I = $$\int {{{dt} \over {{t^7}{{\left( {{1 \over {{t^6}}} + 1} \right)}^{2/3}}}}} $$
Let $${{1 \over {{t^6}}} + 1}$$ = z3
$$ \Rightarrow $$ $$ - {6 \over {{t^7}}}dt = 3{z^2}dz$$
$$ \Rightarrow $$ $${{dt} \over {{t^7}}} = {{{z^2}} \over { - 2}}dz$$
So I = $$ - \int {{{{z^2}dz} \over {2{{\left( {{z^3}} \right)}^{2/3}}}}} $$
I = $$ - \int {{{dz} \over 2}} $$
I = $$ - {z \over 2} + C$$
I = $$ - {1 \over 2}{\left( {1 + {1 \over {{t^6}}}} \right)^{{1 \over 3}}} + C$$
I = $$ - {1 \over 2}{\left( {1 + {1 \over {{{\sin }^6}x}}} \right)^{{1 \over 3}}} + C$$
I = $$ - {1 \over {2{{\sin }^2}x}}{\left( {{{\sin }^6}x + 1} \right)^{{1 \over 3}}} + C$$
$$ \therefore $$ $$\lambda $$ = 3 and f(x) = $$ - {1 \over 2}$$cosec2 x
Then $$\lambda f\left( {{\pi \over 3}} \right)$$ = 3 $$ \times $$ $$ - {1 \over 2}$$cosec2 $${\pi \over 3}$$
= 3 $$ \times $$ $$ - {1 \over 2}$$ $$ \times $$ $${4 \over 3}$$ = -2
Let sin x = t
$$ \Rightarrow $$ cos xdx = dt
$$ \therefore $$ I = $$\int {{{dt} \over {{t^3}{{\left( {1 + {t^6}} \right)}^{2/3}}}}} $$
I = $$\int {{{dt} \over {{t^7}{{\left( {{1 \over {{t^6}}} + 1} \right)}^{2/3}}}}} $$
Let $${{1 \over {{t^6}}} + 1}$$ = z3
$$ \Rightarrow $$ $$ - {6 \over {{t^7}}}dt = 3{z^2}dz$$
$$ \Rightarrow $$ $${{dt} \over {{t^7}}} = {{{z^2}} \over { - 2}}dz$$
So I = $$ - \int {{{{z^2}dz} \over {2{{\left( {{z^3}} \right)}^{2/3}}}}} $$
I = $$ - \int {{{dz} \over 2}} $$
I = $$ - {z \over 2} + C$$
I = $$ - {1 \over 2}{\left( {1 + {1 \over {{t^6}}}} \right)^{{1 \over 3}}} + C$$
I = $$ - {1 \over 2}{\left( {1 + {1 \over {{{\sin }^6}x}}} \right)^{{1 \over 3}}} + C$$
I = $$ - {1 \over {2{{\sin }^2}x}}{\left( {{{\sin }^6}x + 1} \right)^{{1 \over 3}}} + C$$
$$ \therefore $$ $$\lambda $$ = 3 and f(x) = $$ - {1 \over 2}$$cosec2 x
Then $$\lambda f\left( {{\pi \over 3}} \right)$$ = 3 $$ \times $$ $$ - {1 \over 2}$$cosec2 $${\pi \over 3}$$
= 3 $$ \times $$ $$ - {1 \over 2}$$ $$ \times $$ $${4 \over 3}$$ = -2
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