Substitution to Compute an Indefinite Integral with Initial Condition

The question asks the student to find an antiderivative via substitution and then use a given initial condition to determine a specific function value.

jee-main 2025 Q15 View
If $f ( x ) = \int \frac { 1 } { x ^ { 1 / 4 } \left( 1 + x ^ { 1 / 4 } \right) } \mathrm { d } x , f ( 0 ) = - 6$, then $f ( 1 )$ is equal to :
(1) $4 \left( \log _ { e } 2 - 2 \right)$
(2) $2 - \log _ { e ^ { 2 } } 2$
(3) $\log _ { e } 2 + 2$
(4) $4 \left( \log _ { e } 2 + 2 \right)$
jee-main 2025 Q74 View
Q74. If $\int \frac { 1 } { \mathrm { a } ^ { 2 } \sin ^ { 2 } x + \mathrm { b } ^ { 2 } \cos ^ { 2 } x } \mathrm {~d} x = \frac { 1 } { 12 } \tan ^ { - 1 } ( 3 \tan x ) +$ constant, then the maximum value of $\mathrm { a } \sin x + \mathrm { b } \cos x$, is :
(1) $\sqrt { 40 }$
(2) $\sqrt { 41 }$
(3) $\sqrt { 39 }$
(4) $\sqrt { 42 }$
jee-main 2026 Q29 View
If $F ( t ) = \int \frac { 1 - \sin ( \ln t ) } { 1 - \cos ( \ln t ) } d t$ and $F \left( e ^ { \pi / 2 } \right) = - e ^ { \pi / 2 }$ then $F \left( e ^ { \pi / 4 } \right)$ is:
(A) $( - 1 - \sqrt { 2 } ) \mathrm { e } ^ { \frac { \pi } { 4 } }$
(B) $( 1 - \sqrt { 2 } ) e ^ { \frac { \pi } { 4 } }$
(D) $( - 2 - \sqrt { 2 } ) e ^ { \frac { \pi } { 4 } }$
tmua 2019 Q12 1 marks View
It is given that
$$\frac{dV}{dt} = \frac{24\pi(t-1)}{(1+\sqrt{t})} \text{ for } t \geq 1$$
and $V = 7$ when $t = 1$. Find the value of $V$ when $t = 9$.