Q.Evaluate $\displaystyle\int x\cos x\,dx$.

Showing the 12 most recent of 40 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\int\log x\,dx =$

   (a) $\frac{1}{x} + k$

   (b) $x\log x + k$

   (c) $x\log x - x + k$

   (d) $x\log x + x + k$

   ›Reveal solutionSolution

   Integrate by parts: $\int\log x\,dx=x\log x-x+k$.

   Take $u=\log x$ and $dv=dx$, so $du=\tfrac{1}{x}\,dx$ and $v=x$:

   $\int\log x\,dx=x\log x-\int x\cdot\tfrac{1}{x}\,dx=x\log x-\int 1\,dx=x\log x-x+k$.

   ✓Final answer

   (C) $x\log x - x + k$.

2. CBSE 2026Set A1 markMCQ
   Q.$\int\cos\sqrt{x}\,dx =$

   (a) $\sin\sqrt{x} + \cos\sqrt{x} + k$

   (b) $\frac{1}{2}(\sqrt{x}\sin\sqrt{x} - \cos\sqrt{x}) + k$

   (c) $2(\sqrt{x}\sin\sqrt{x} + \cos\sqrt{x}) + k$

   (d) $\sin\sqrt{x} + k$

   ›Reveal solutionSolution

   Put $t=\sqrt{x}$; the integral becomes $2\int t\cos t\,dt=2(t\sin t+\cos t)+k$.

   Let $t=\sqrt{x}$, so $x=t^2$ and $dx=2t\,dt$. Then

   $\int\cos\sqrt{x}\,dx=\int\cos t\,(2t\,dt)=2\int t\cos t\,dt$.

   Integrate $\int t\cos t\,dt$ by parts ($u=t,\ dv=\cos t\,dt$): $=t\sin t-\int\sin t\,dt=t\sin t+\cos t$.

   Hence the answer is $2(t\sin t+\cos t)+k=2(\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x})+k$.

   ✓Final answer

   (C) $2(\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x})+k$.

3. CBSE 2026Set A1 markMCQ
   Q.$\int e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)dx =$

   (a) $e^x\tan^{-1}x + k$

   (b) $e^x \cdot \frac{1}{1+x^2} + k$

   (c) $e^x + k$

   (d) $\tan^{-1}x + k$

   ›Reveal solutionSolution

   Recognise $\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+k$; here $f(x)=\tan^{-1}x$.

   Note $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$. So the integrand is $e^x\big[\tan^{-1}x+(\tan^{-1}x)'\big]$, which matches the standard pattern

   $\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+k$.

   Therefore the integral equals $e^x\tan^{-1}x+k$.

   ✓Final answer

   (A) $e^x\tan^{-1}x+k$.

4. CBSE 2026Set A1 markMCQ
   Q.$\int_0^1 x e^x\,dx =$

   (a) $1$

   (b) $0$

   (c) $2$

   (d) $-1$

   ›Reveal solutionSolution

   By parts: $\int_0^1 xe^x\,dx=[(x-1)e^x]_0^1=1$.

   Take $u=x,\ dv=e^x\,dx$, so $du=dx,\ v=e^x$:

   $\int x e^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x=(x-1)e^x$.

   Evaluate from $0$ to $1$: $(1-1)e^1-(0-1)e^0=0-(-1)=1$.

   ✓Final answer

   (A) $1$.

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the value of $\int e^x(\sin x-\cos x)\,dx$.

   (a) $-e^x\cos x+c$

   (b) $e^x\sin x+c$

   (c) $-e^x\sec x+c$

   (d) $e^x\operatorname{cosec} x+c$

   ›Reveal solutionSolution

   Recognising the standard form $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+c$ gives $-e^x\cos x+c$.

   There is a standard integration result:

   $\int e^x\big[f(x)+f'(x)\big]\,dx=e^x f(x)+c$

   Compare the integrand $\sin x-\cos x$ with $f(x)+f'(x)$. Try $f(x)=-\cos x$; then $f'(x)=\sin x$, so

   $f(x)+f'(x)=-\cos x+\sin x=\sin x-\cos x$

   which exactly matches the given integrand.

   Therefore

   $\int e^x(\sin x-\cos x)\,dx=e^x\cdot(-\cos x)+c=-e^x\cos x+c$

   ✓Final answer

   The correct option is (a) $-e^x\cos x+c$.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\int e^x (\log \sec x + \tan x)\, dx =$

   (a) $e^x + C$

   (b) $e^x \tan x + C$

   (c) $e^x(\log \sec x) + C$

   (d) None of these

   ›Reveal solutionSolution

   This is of the standard form $\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C$.

   Let $f(x) = \log\sec x$. Then $f'(x) = \dfrac{\sec x\tan x}{\sec x} = \tan x$.

   So the integrand $e^x(\log\sec x + \tan x) = e^x[f(x)+f'(x)]$, and by the standard result:

   $\int e^x(\log\sec x+\tan x)\,dx = e^x \log\sec x + C$.

   ✓Final answer

   (c) $e^x(\log\sec x)+C$.

7. CBSE 2025Set X11 markMCQ
   Q.$\int e^x (\sin x - \cos x)\, dx$ is

   (a) $-e^x \cos x$

   (b) $e^x \cos x$

   (c) $e^x \sin x$

   (d) $e^x \sin^2 x$

   ›Reveal solutionSolution

   Integral of the form $\int e^x(f+f')\,dx = e^x f$ — correct option (a).

   Take $f(x) = -\cos x$, so $f'(x) = \sin x$. Then $f + f' = -\cos x + \sin x = \sin x - \cos x$, exactly the integrand. Hence $\int e^x(\sin x - \cos x)\,dx = e^x f(x) = -e^x\cos x + C$.

   ✓Final answer

   (a) $-e^x \cos x$

8. CBSE 2025Set ANNUAL1 mark
   Q.Find $\displaystyle\int x \cdot e^x\,dx$.

   ›Reveal solutionSolution

   Apply integration by parts with $u=x$, $dv=e^x dx$.

   $\int x\,e^x\,dx = x\,e^x - \int e^x\,dx = x\,e^x - e^x + c = e^x(x-1)+c$

   ✓Final answer

   $e^x(x-1) + c$

9. CBSE 2025Set E1 markMCQ
   Q.$\int \log x^2\,dx =$

   (a) $\frac{1}{x^2} + k$

   (b) $\frac{2}{x} + k$

   (c) $x\log x - x + k$

   (d) $2(x\log x - x) + k$

   ›Reveal solutionSolution

   Bring down the power, then integrate $\log x$ by parts; result $2(x\log x-x)+k$.

   First $\log x^2=2\log x$. Now integrate $\int\log x\,dx$ by parts with $u=\log x$, $dv=dx$:

   $\int\log x\,dx=x\log x-\int x\cdot\frac1x\,dx=x\log x-x.$

   Therefore

   $\int\log x^2\,dx=2\int\log x\,dx=2(x\log x-x)+k.$

   ✓Final answer

   (D) $2(x\log x-x)+k$.

10. CBSE 2025Set ANNUAL1 mark
    Q.Evaluate $\displaystyle\int x\sin x\,dx$.

    ›Reveal solutionSolution

    Use integration by parts (ILATE): take $u=x$ (algebraic) and $dv=\sin x\,dx$.

    Let $u=x$, $dv=\sin x\,dx$, so $du=dx$, $v=-\cos x$.

    $\int x\sin x\,dx=uv-\int v\,du=-x\cos x-\int(-\cos x)\,dx=-x\cos x+\int\cos x\,dx$

    $=-x\cos x+\sin x+C$

    ✓Final answer

    $-x\cos x+\sin x+C$.

11. CBSE 2025Set ANNUAL1 mark
    Q.Evaluate : $\displaystyle\int x\log 2x\, dx$

    ›Reveal solutionSolution

    Use integration by parts, taking the logarithm as the first function.

    Let $I=\displaystyle\int x\log 2x\,dx$. Apply $\int u\,dv=uv-\int v\,du$ with

    $u=\log 2x,\qquad dv=x\,dx.$

    Then

    $du=\frac{1}{2x}\cdot 2\,dx=\frac1x\,dx,\qquad v=\frac{x^2}{2}.$

    Therefore

    $I=\log 2x\cdot\frac{x^2}{2}-\int\frac{x^2}{2}\cdot\frac1x\,dx=\frac{x^2}{2}\log 2x-\frac12\int x\,dx.$

    And $\displaystyle\frac12\int x\,dx=\frac12\cdot\frac{x^2}{2}=\frac{x^2}{4}$, so

    $I=\frac{x^2}{2}\log 2x-\frac{x^2}{4}+C.$

    ✓Final answer

    $\displaystyle\int x\log 2x\,dx=\frac{x^2}{2}\log 2x-\frac{x^2}{4}+C$

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\displaystyle\int e^x\sec x(1+\tan x)\, dx$ is equal to

    (a) $e^x\cos x + c$

    (b) $e^x\sec x + c$

    (c) $e^x\sin x + c$

    (d) $e^x\tan x + c$

    ›Reveal solutionSolution

    Recognise the form $\int e^x\{f(x)+f'(x)\}\,dx=e^xf(x)+c$.

    Expand the integrand:

    $e^x\sec x(1+\tan x)=e^x\big(\sec x+\sec x\tan x\big).$

    Let $f(x)=\sec x$. Then

    $f'(x)=\sec x\tan x.$

    So the integrand is exactly $e^x\{f(x)+f'(x)\}$, and by the standard result

    $\int e^x\{f(x)+f'(x)\}\,dx=e^xf(x)+c=e^x\sec x+c.$

    Verification (differentiate the answer):

    $\frac{d}{dx}\big(e^x\sec x\big)=e^x\sec x+e^x\sec x\tan x=e^x\sec x(1+\tan x),$

    which is the original integrand. This confirms option (b).

    ✓Final answer

    (b) $e^x\sec x+c$