Q.Find $\displaystyle\int \dfrac{\sin(x-a)}{\sin(x+a)}\, dx$.

Showing the 12 most recent of 103 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.One of the values of $x$ for which $\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = 1$ is (A) $0$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   The determinant simplifies to $\sin x \cos x + \sin x \cos x = \sin 2x$. Setting $\sin 2x = 1$ gives $2x = \frac{\pi}{2} + 2n\pi$, so $x = \frac{\pi}{4}$ is one solution. The correct option is (B).

   The problem gives a $2 \times 2$ determinant equal to $1$ and asks for a value of $x$ from the options. The fastest route is to compute the determinant directly — it’s a simple expression in $\sin x$ and $\cos x$ — and then solve the resulting trigonometric equation.

   The determinant of $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad - bc$. Here:

   $\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = (\cos x)(\sin x) - (\sin x)(-\cos x)$

   1. Simplify the expression. The first term is $\cos x \sin x$. The second term: $(\sin x)(-\cos x) = -\sin x \cos x$, but there’s a minus sign in front, so it becomes $-(-\sin x \cos x) = +\sin x \cos x$. So the determinant equals:

   $\cos x \sin x + \sin x \cos x = 2 \sin x \cos x$

   2. Use the double-angle identity. Recall that $2 \sin x \cos x = \sin 2x$. Therefore the equation becomes:

   $\sin 2x = 1$

   3. Solve $\sin 2x = 1$. The sine function equals $1$ at $\frac{\pi}{2}$ plus any integer multiple of $2\pi$:

   $2x = \frac{\pi}{2} + 2n\pi \quad \Rightarrow \quad x = \frac{\pi}{4} + n\pi$

   where $n$ is any integer.

   4. Check the given options.
      • (A) $0$: $\sin 0 = 0$, not $1$.
      • (B) $\frac{\pi}{4}$: $\sin \frac{\pi}{2} = 1$ — works. …

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2. CBSE 2026Set CX1 markMCQ
   Q.$\sin(\tan^{-1}x)$, $|x|<1$ is equal to:

   (a) $\dfrac{x}{\sqrt{1+x^{2}}}$

   (b) $\dfrac{x}{\sqrt{1-x^{2}}}$

   (c) $\dfrac{1}{\sqrt{1+x^{2}}}$

   (d) $\dfrac{1}{\sqrt{1-x^{2}}}$

   ›Reveal solutionSolution

   With $\theta=\tan^{-1}x$ a right triangle gives $\sin\theta=\dfrac{x}{\sqrt{1+x^{2}}}$ — option (a).

   Concept: Convert the inverse function to an angle and read the ratio off a right triangle.

   Let $\theta=\tan^{-1}x$, so $\tan\theta=x$. Take the opposite side $=x$ and adjacent $=1$; then the hypotenuse is $\sqrt{1+x^{2}}$.

   …

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3. CBSE 2026Set A1 markMCQ
   Q.$\sin(\cos^{-1}3/5) =$

   (a) $\frac{3}{4}$

   (b) $\frac{4}{5}$

   (c) $\frac{3}{5}$

   (d) $\frac{5}{4}$

   ›Reveal solutionSolution

   $\sin\left(\cos^{-1}\frac{3}{5}\right) = \frac{4}{5}$.

   Let $\theta = \cos^{-1}\frac{3}{5}$, so $\cos\theta = \frac{3}{5}$ with $\theta\in[0,\pi]$, where $\sin\theta \ge 0$.

   Then …

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4. CBSE 2026Set A1 markMCQ
   Q.If $|x| \le 1$, then $\tan(\cos^{-1}x) =$

   (a) $\frac{\sqrt{1-x^2}}{x}$

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

   (c) $\frac{\sqrt{1+x^2}}{x}$

   (d) $\sqrt{1-x^2}$

   ›Reveal solutionSolution

   $\tan(\cos^{-1}x) = \frac{\sqrt{1-x^2}}{x}$.

   Let $\theta = \cos^{-1}x$, so $\cos\theta = x$ with $\theta\in[0,\pi]$ (where $\sin\theta\ge 0$).

   Then $\sin\theta = \sqrt{1-x^2}$, and …

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5. CBSE 2026Set A1 markMCQ
   Q.$\int\frac{\cos 2x}{(\sin x + \cos x)^2}\,dx =$

   (a) $2\log(\sin x + \cos x) + k$

   (b) $\log(\sin x + \cos x) + k$

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

   (d) $-\frac{1}{\sin x + \cos x} + k$

   ›Reveal solutionSolution

   Factor $\cos 2x$ and substitute $u=\sin x+\cos x$ to get $\log(\sin x+\cos x)+k$.

   Write $\cos 2x=\cos^2 x-\sin^2 x=(\cos x-\sin x)(\cos x+\sin x)$. Then

   $\dfrac{\cos 2x}{(\sin x+\cos x)^2}=\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2}=\dfrac{\cos x-\sin x}{\sin x+\cos x}$.

   …

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6. CBSE 2026Set A1 markMCQ
   Q.$\int\frac{1 - \cos 2x}{1 + \cos 2x}\,dx =$

   (a) $\tan x + x + k$

   (b) $\tan x - x + k$

   (c) $x - \tan^2 x + k$

   (d) $\tan\frac{x}{2} + k$

   ›Reveal solutionSolution

   Simplify to $\tan^2 x$, then integrate: $\tan x - x + k$.

   Use $1-\cos 2x=2\sin^2 x$ and $1+\cos 2x=2\cos^2 x$:

   $\dfrac{1-\cos 2x}{1+\cos 2x}=\dfrac{2\sin^2 x}{2\cos^2 x}=\tan^2 x=\sec^2 x-1$.

   …

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7. 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 $v=x$:

   …

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

   …

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9. 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

   …

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10. 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 $du=dx,\ v=e^x$:

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

    …

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11. 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

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

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12. CBSE 2026Set ANNUAL1 markMCQ
    Q.$\int \frac{dx}{\sin^2 x \cos^2 x}$ equals

    (a) $\tan x + \sin x + c$

    (b) $\tan x - \cot x + c$

    (c) $\tan x \cot x + c$

    (d) $2\tan x - \cot 2x + c$

    ›Reveal solutionSolution

    Split the integrand using $\sin^2x+\cos^2x=1$ in the numerator, then integrate each standard term.

    $\dfrac{1}{\sin^2x\cos^2x} = \dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} = \dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x} = \sec^2x+\csc^2x$.

    …

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