Q.Integrate the following function: $\sin 2x$

Showing the 12 most recent of 17 on this concept.

1. CBSE 20241 markMCQ
   Q.$\int \frac{\cos 2x}{\sin^2 x \cos^2 x} dx$ is equal to : (A) $\cot x + \tan x + c$ (B) $-\cot x + \tan x + c$ (C) $\cot x - \tan x + c$ (D) $-\cot x - \tan x + c$

   ›Reveal solutionSolution

   The integral simplifies by rewriting $\cos 2x$ as $\cos^2 x - \sin^2 x$, splitting into two simple integrals, and integrating term by term. The result is $-\cot x - \tan x + c$, which matches option (D).

   The key insight here is that the denominator $\sin^2 x \cos^2 x$ is a product of squares, and the numerator $\cos 2x$ is begging to be expressed in terms of $\sin^2 x$ and $\cos^2 x$. Once you do that, the fraction splits naturally into two separate terms, each of which is a standard integral.

   Let’s walk through it.

   1. Rewrite $\cos 2x$ using a double-angle identity.

      The most useful form here is $\cos 2x = \cos^2 x - \sin^2 x$. This directly matches the squares in the denominator.

   2. Split the integrand into two fractions.

   $\frac{\cos 2x}{\sin^2 x \cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} = \frac{\cos^2 x}{\sin^2 x \cos^2 x} - \frac{\sin^2 x}{\sin^2 x \cos^2 x}$

   Cancel common factors:

   $= \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x}$

   3. Recognise the standard forms.

   $\frac{1}{\sin^2 x} = \csc^2 x, \quad \frac{1}{\cos^2 x} = \sec^2 x$

   So the integral becomes:

   $\int (\csc^2 x - \sec^2 x) \, dx$

   4. Integrate term by term. Recall:

   $\int \csc^2 x \, dx = -\cot x + c, \quad \int \sec^2 x \, dx = \tan x + c$

   Therefore:

   $\int (\csc^2 x - \sec^2 x) \, dx = -\cot x - \tan x + c$

   Watch out

   A common mistake is to forget the minus sign when integrating $\csc^2 x$. The derivative of $\cot x$ is $-\csc^2 x$, so the antiderivative of $\csc^2 x$ is $-\cot x$, not $\cot x$.

   Tip

   If you ever see $\sin^2 x \cos^2 x$ in the denominator and $\cos 2x$ in the numerator, this splitting trick works every time. Alternatively, you could use $\cos 2x = 1 - 2\sin^2 x$ or $2\cos^2 x - 1$, but the symmetric form $\cos^2 x - \sin^2 x$ is the cleanest here.

   ✓Final answer

   The correct option is (D): $-\cot x - \tan x + c$.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \sqrt{1+\cos 2x}\, dx =$

   (a) 0

   (b) 1

   (c) $\dfrac{1}{\sqrt{2}}$

   (d) None of these

   ›Reveal solutionSolution

   Use $1+\cos 2x = 2\cos^2 x$ to simplify the square root, then integrate $\cos x$.

   Since $1+\cos 2x = 2\cos^2 x$, we have $\sqrt{1+\cos 2x} = \sqrt{2}\,|\cos x|$. On $[0,\pi/2]$, $\cos x \ge 0$, so this is $\sqrt{2}\cos x$.

   $\displaystyle\int_0^{\pi/2}\sqrt{1+\cos 2x}\,dx = \sqrt{2}\int_0^{\pi/2}\cos x\,dx = \sqrt{2}\Big[\sin x\Big]_0^{\pi/2} = \sqrt{2}(1-0) = \sqrt{2}$

   This value ($\sqrt2 \approx 1.414$) is not $0$, $1$, or $\dfrac{1}{\sqrt2}$.

   ✓Final answer

   (d) None of these — the correct value is $\sqrt{2}$.

3. CBSE 2025Set ANNUAL1 markMCQ
   Q.Value of $\displaystyle\int \cos^2 x\, dx$ is:

   (a) $\dfrac14 \sin 2x + c$

   (b) $\dfrac{x}{2} + \dfrac14 \sin 2x + c$

   (c) $\sin^2 x + c$

   (d) $\dfrac{x}{2} - \dfrac14 \cos 2x + c$

   ›Reveal solutionSolution

   Use the identity $\cos^2 x = \dfrac{1+\cos 2x}{2}$.

   $\int \cos^2 x\,dx = \int \dfrac{1+\cos 2x}{2}\,dx = \dfrac{x}{2}+\dfrac14\sin 2x + c$

   ✓Final answer

   (ii) $\dfrac{x}{2} + \dfrac14 \sin 2x + c$

4. CBSE 2025Set E1 markMCQ
   Q.$\int_{0}^{\pi/2} \sin x \cdot \cos x\,dx =$

   (a) $1$

   (b) $\frac{1}{2}$

   (c) $-1$

   (d) $\frac{1}{4}$

   ›Reveal solutionSolution

   With $u=\sin x$, $\int_0^{\pi/2}\sin x\cos x\,dx=\big[\tfrac{\sin^2x}{2}\big]_0^{\pi/2}=\tfrac12$.

   Let $u=\sin x$, so $du=\cos x\,dx$. Then

   $\int_0^{\pi/2}\sin x\cos x\,dx = \int_0^{1} u\,du = \Big[\frac{u^2}{2}\Big]_0^1 = \frac{1}{2}.$

   ✓Final answer

   (B) $\frac{1}{2}$.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\int_{0}^{\pi/4} \sin 2x\,dx$ is equal to

   (a) 0

   (b) 1

   (c) 2

   (d) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Antiderivative of $\sin2x$ is $-\cos2x/2$; evaluating from $0$ to $\pi/4$ gives $1/2$.

   $\int_{0}^{\pi/4}\sin2x\,dx=\left[-\dfrac{\cos2x}{2}\right]_{0}^{\pi/4}$

   $=-\dfrac{\cos(\pi/2)}{2}-\left(-\dfrac{\cos0}{2}\right)=-\dfrac{0}{2}+\dfrac{1}{2}=\dfrac{1}{2}$

   ✓Final answer

   (d) $\dfrac{1}{2}$

6. CBSE 2024Set D1 markMCQ
   Q.$\int_0^{\pi/2} \cos 2x\,dx =$

   (a) $0$

   (b) $1$

   (c) $-1$

   (d) $2$

   ›Reveal solutionSolution

   The antiderivative $\frac{\sin 2x}{2}$ evaluates to $0$ at both limits.

   $\displaystyle\int_0^{\pi/2}\cos 2x\,dx = \left[\dfrac{\sin 2x}{2}\right]_0^{\pi/2} = \dfrac{\sin\pi}{2} - \dfrac{\sin 0}{2} = 0 - 0 = 0$.

   ✓Final answer

   (a) $0$.

7. CBSE 2024Set ANNUAL1 markMCQ
   Q.$\int \sin^2 x \,dx =$

   (a) $\frac{x}{2}-\frac{\sin 2x}{4}+c$

   (b) $\frac{\sin 2x}{4}-\frac{x}{2}+c$

   (c) $\frac{\sin 2x}{2}+\frac{x}{2}+c$

   (d) none of these

   ›Reveal solutionSolution

   Rewrite sin^2 x using the double-angle identity, then integrate term by term.

   Using $\sin^2x=\dfrac{1-\cos2x}{2}$:

   $\int\sin^2x\,dx=\int\dfrac{1-\cos2x}{2}\,dx=\dfrac{x}{2}-\dfrac{\sin2x}{4}+c$

   ✓Final answer

   (a) $\dfrac{x}{2}-\dfrac{\sin2x}{4}+c$.

8. CBSE 2023Set AX1 markMCQ
   Q.The value of $\int \cos^2 x\, dx$ will be

   (a) $\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+c$

   (b) $\dfrac{x}{4}-\dfrac{1}{2}\sin 2x+c$

   (c) $\cos^2 x-\sin^2 x+c$

   (d) $2\cos x\sin x+\dfrac{x}{2}+c$

   ›Reveal solutionSolution

   Reduce the square with the double-angle identity, then integrate term by term: $\dfrac{x}{2}+\dfrac14\sin 2x+c$, option (a).

   Concept. A squared trig function is integrated by lowering the power with a double-angle identity, turning $\cos^2x$ into a simple linear combination of $1$ and $\cos 2x$.

   Solution. Using $\cos^2x=\dfrac{1+\cos 2x}{2}$,

   $\int\cos^2x\,dx=\int\frac{1+\cos 2x}{2}\,dx=\frac12\int dx+\frac12\int\cos 2x\,dx.$

   $=\frac{x}{2}+\frac12\cdot\frac{\sin 2x}{2}+c=\frac{x}{2}+\frac{1}{4}\sin 2x+c.$

   ✓Final answer

   (a) $\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+c$

9. CBSE 2023Set ANNUAL1 markMCQ
   Q.The value of $\int \cos^2 x\,dx$ is:

   (a) $\frac{x}{2}+\frac{1}{4}\sin 2x+C$

   (b) $x^2+\frac{1}{4}\sin 2x+C$

   (c) $\frac{x}{4}+\frac{1}{2}\sin x+C$

   (d) $\frac{x^2}{2}+\frac{1}{2}\sin^2 x+C$

   ›Reveal solutionSolution

   Use the identity $\cos^2 x=\frac{1+\cos 2x}{2}$ to make the integral straightforward.

   $\int\cos^2x\,dx=\int\dfrac{1+\cos 2x}{2}dx=\dfrac{1}{2}\int dx+\dfrac{1}{2}\int\cos 2x\,dx$

   $=\dfrac{x}{2}+\dfrac{1}{2}\cdot\dfrac{\sin 2x}{2}+C=\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+C$

   ✓Final answer

   (a) $\frac{x}{2}+\frac{1}{4}\sin 2x+C$.

10. CBSE 2023Set ANNUAL1 mark
    Q.Evaluate : $\displaystyle\int_0^{\pi/2} \cos 2x\, dx$ OR Evaluate : $\displaystyle\int_0^1 \dfrac{dx}{1+x^2}$

    ›Reveal solutionSolution

    Integrate $\cos2x$ to $\tfrac12\sin2x$ and evaluate the limits.

    $\int_0^{\pi/2}\cos 2x\,dx=\left[\frac{\sin 2x}{2}\right]_0^{\pi/2}=\frac{\sin\pi}{2}-\frac{\sin 0}{2}=\frac{0}{2}-\frac{0}{2}=0.$

    ✓Final answer

    $\displaystyle\int_0^{\pi/2}\cos 2x\,dx=0$.

    Alternative (Or):

    The integrand is the derivative of $\tan^{-1}x$.

    $\int_0^1\frac{dx}{1+x^2}=\big[\tan^{-1}x\big]_0^1=\tan^{-1}1-\tan^{-1}0=\frac{\pi}{4}-0=\frac{\pi}{4}.$

    ✓Final answer

    $\dfrac{\pi}{4}$.

11. CBSE 2022Set ANNUAL1 markMCQ
    Q.$\int_{\pi/2}^{\pi/2} \sin^2 x\,dx =$

    (a) $0$

    (b) $1$

    (c) $5$

    (d) $11$

    ›Reveal solutionSolution

    The limits of integration are identical, so the integral is $0$.

    The printed integral is $\int_{\pi/2}^{\pi/2}\sin^2 x\,dx$. By the property $\int_a^a f(x)\,dx = 0$, any definite integral whose lower and upper limits coincide equals zero.

    $\int_{\pi/2}^{\pi/2}\sin^2 x\,dx = 0$.

    ✓Final answer

    (a) $0$.

12. CBSE 2021Set I1 markMCQ
    Q.$\int_0^{\pi/2} \cos x\,dx =$

    (a) $1$

    (b) $-1$

    (c) $0$

    (d) $2$

    ›Reveal solutionSolution

    Antiderivative of $\cos x$ is $\sin x$; evaluate at the limits.

    $\int_0^{\pi/2}\cos x\,dx=[\sin x]_0^{\pi/2}=\sin\dfrac{\pi}{2}-\sin 0=1-0=1$.

    ✓Final answer

    (a) $1$.