JEE MAIN - Mathematics (2013 (Offline) - No. 7)
If $$\int {f\left( x \right)dx = \psi \left( x \right),} $$ then $$\int {{x^5}f\left( {{x^3}} \right)dx} $$ is equal to
$${1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C$$
$${1 \over 3}{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^3}\psi \left( {{x^3}} \right)dx} + C$$
$${1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} + C$$
$${1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^3}\psi \left( {{x^3}} \right)dx} } \right] + C$$
Explanation
Let $$\int {f\left( x \right)dx = \psi \left( x \right)} $$
Let $$I = \int {{x^5}} f\left( {{x^3}} \right)dx$$
put $${x^3} = t \Rightarrow 3{x^2}dx = dt$$
$$I = {1 \over 3}\int {3.{x^2}} .{x^3}.f\left( {{x^3}} \right).dx$$
$$ = {1 \over 3}\int {tf} \left( t \right)dt$$
$$ = {1 \over 3}\left[ {t\int {f\left( t \right)dt - \int {f\left( t \right)dt} } } \right]$$
$$ = {1 \over 3}\left[ {t\psi \left( t \right) - \int {\psi \left( t \right)dt} } \right]$$
$$ = {1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C$$
$$ = {1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}} \psi \left( {{x^3}} \right)dx + C$$
Let $$I = \int {{x^5}} f\left( {{x^3}} \right)dx$$
put $${x^3} = t \Rightarrow 3{x^2}dx = dt$$
$$I = {1 \over 3}\int {3.{x^2}} .{x^3}.f\left( {{x^3}} \right).dx$$
$$ = {1 \over 3}\int {tf} \left( t \right)dt$$
$$ = {1 \over 3}\left[ {t\int {f\left( t \right)dt - \int {f\left( t \right)dt} } } \right]$$
$$ = {1 \over 3}\left[ {t\psi \left( t \right) - \int {\psi \left( t \right)dt} } \right]$$
$$ = {1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C$$
$$ = {1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}} \psi \left( {{x^3}} \right)dx + C$$
Comments (0)
