MCQ
$\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$ is equal to
- ✓$\tan \left(x e^x\right)+c$
- B$\cot \left(x e^x\right)+c$
- C$\cot \left(e^x\right)+c$
- D$\tan \left[e^x(1+x)\right]+c$
✓
Answer
Correct option: A.
$\tan \left(x e^x\right)+c$
$\text {Let } I=\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x$
$\text { Put } x e^x=t$
$\Rightarrow\left(x e^x+e^x\right) d x=d t$
$\Rightarrow e^x(x+1) d x=d t$
$\therefore I=\int \frac{d t}{\cos ^2 t}=\int \sec ^2 t d t=\tan t+c$
$=\tan \left(x e^x\right)+c$
$\text { Put } x e^x=t$
$\Rightarrow\left(x e^x+e^x\right) d x=d t$
$\Rightarrow e^x(x+1) d x=d t$
$\therefore I=\int \frac{d t}{\cos ^2 t}=\int \sec ^2 t d t=\tan t+c$
$=\tan \left(x e^x\right)+c$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A cylindrical vessel of radius 0.5m is filled with oil at the rate of $0.25\pi/\text{minute}$. The rate at which the surface of the oil is rising, is:
→If two constraints do not intersect in the positive quadrant of the graph, then.
The area bounded by $y=x e^{|x|}$ and $|x|=1$ is:
→If $d$ is the determinant of a square matrix $A$ of order $n,$ then the determinant of its adjoint is:
→$y = 4\sin 3x$ is a solution of the differential equation
→If $f(x) = {\cot ^{ - 1}}\left( {{{{x^x} - {x^{ - x}}} \over 2}} \right)\,,$ then $f'(1)$ is equal to
→Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function $Z=3 x+9 y$ maximum?
→Find the values of $x, y, z$ and $w$ respectively such that $\left[\begin{array}{cc}x-y & 2 z+w \\ 2 x-y & 2 x+w\end{array}\right]=\left[\begin{array}{cc}5 & 3 \\ 12 & 15\end{array}\right]$.
→Choose the correct answer from the given four options. The feasible solution for a LPP is shown in. Let Z = 3x - 4y be the objective function.
Maximum of Z occurs at:
Maximum of Z occurs at:
$\int_{ - a}^a {\sin x\,f(\cos x)\,dx = } $
→