Anand Classes offers well-organized NCERT Solutions for Integrals Exercise 7.10 of Chapter 7 for Class 12 Mathematics, designed to help students easily understand complex integration concepts with step-by-step solutions. This Set-2 PDF includes solved examples, concept explanations, and exam-focused notes based on the latest NCERT curriculum, making it ideal for revision and board exam preparation. Click the print button to download study material and notes.
NCERT Question.11 : Evaluate the integral
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{2}x\;dx$$
Solution
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{2}x\;dx$$
Recognize symmetry.
Since $\sin^{2}x$ is an even function,
$$\sin^{2}(-x)=\sin^{2}x,$$
so
$$I=2\int_{0}^{\frac{\pi}{2}}\sin^{2}x\;dx.$$
Use the identity
$$\sin^{2}x=\frac{1-\cos2x}{2}.$$
Thus,
$$I=2\int_{0}^{\frac{\pi}{2}}\frac{1-\cos2x}{2}\;dx
=\int_{0}^{\frac{\pi}{2}}\bigl(1-\cos2x\bigr)\;dx.$$
Integrate termwise:
$$
I =\int_{0}^{\frac{\pi}{2}}1\;dx-\int_{0}^{\frac{\pi}{2}}\cos2x\;dx $$
$$=\Big[x\Big]_{0}^{\frac{\pi}{2}}-\Big[\tfrac{1}{2}\sin2x\Big]_{0}^{\frac{\pi}{2}}$$
$$=\frac{\pi}{2}-\tfrac{1}{2}(\sin\pi-\sin0)$$
$$=\frac{\pi}{2}-0.$$
Final Result
$$\boxed{\;I=\frac{\pi}{2}\;}$$
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NCERT Question.12 : Evaluate the integral
$$I=\int_{0}^{\pi}\frac{x}{1+\sin x}\;dx$$
Solution :
Let
$$I=\int_{0}^{\pi}\frac{x}{1+\sin x}\;dx$$
Step 1: Use the property
using the integration property
$$I = \int_0^{\pi} f(x)dx = \int_0^{\pi} f(\pi – x)dx$$
Apply the substitution
$$
x \;\to\; \pi – x.
$$
Then
$$
I=\int_{0}^{\pi}\frac{\pi – x}{1+\sin(\pi – x)}\;dx
$$
But
$$
\sin(\pi – x)=\sin x\;
$$
so
$$
I=\int_{0}^{\pi}\frac{\pi – x}{1+\sin x}\;dx
$$
Step 2: Add the two integrals
$$
2I=\int_{0}^{\pi}\frac{x}{1+\sin x}\;dx + \int_{0}^{\pi}\frac{\pi – x}{1+\sin x}\;dx
$$
$$
2I=\int_{0}^{\pi}\frac{\pi}{1+\sin x}\;dx
$$
$$
2I=\pi\int_{0}^{\pi}\frac{dx}{1+\sin x}
$$
Let
$$
J=\int_{0}^{\pi}\frac{dx}{1+\sin x}.
$$
Then
$$
2I=\pi J.
$$
Step 3: Evaluate
$$ J=\int_{0}^{\pi}\frac{dx}{1+\sin x}.$$
Use the substitution
$$
t=\tan\frac{x}{2}\; =>\qquad x=2\tan^{-1} t
$$
$$
\sin x = \frac{2t}{1+t^2}\; =>\qquad dx=\frac{2\;dt}{1+t^2}
$$
Then
$$
J=\int_{0}^{\infty}\frac{1+t^2}{1+t^2+2t}\cdot \frac{2\;dt}{1+t^2}
$$
$$
J=\int_{0}^{\infty}\frac{2\;dt}{(t+1)^2}
$$
$$
J=2\int_{0}^{\infty}\frac{dt}{(t+1)^2}
$$
$$
J=2\left[\frac{-1}{t+1}\right]_{0}^{\infty}
$$
$$
J=2(0 – (-1)) = 2
$$
Thus
$$
J=2.
$$
Step 4: Final value
$$
2I=\pi J = \pi \cdot 2 = 2\pi
$$
$$
I=\pi
$$
Final Answer
$$\boxed{\;\pi\;}$$
NCERT Question.13 : Evaluate the integral
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{7}x\;dx$$
Solution
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{7}x\;dx$$
Observation โ parity of the integrand.
The sine function is odd: $\sin(-x)=-\sin x$. An odd power of an odd function is odd, so
$$\sin^{7}(-x)=\big(\sin(-x)\big)^{7}=(-\sin x)^{7}=-\sin^{7}x.$$
Thus $\sin^{7}x$ is an odd function.
Key property (symmetry).
The integral of an odd function over symmetric limits $[-a,a]$ is zero:
$$\int_{-a}^{a} \text{(odd function)}dx=0.$$
Apply this with $a=\dfrac{\pi}{2}$:
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^{7}x\;dx=0.$$
Final Result
$$\boxed{\;I=0\;}$$
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NCERT Question.13 : Evaluate the integral
$$I=\int_{0}^{2\pi}\cos^{5}x\;dx$$
Solution :
use the odd-power cosine identity:
$$\cos^{5}x=\cos^{4}x\cos x=(\cos^{2}x)^2\cos x=(1-\sin^{2}x)^{2}\cos x.$$
Let
$$
t=\sin x \quad\Rightarrow\quad dt=\cos x\;dx.
$$
Also note how $(t=\sin x)$ changes over one full period:
- When $(x=0):โ(t=0)$
- When $(x=\pi):โ(t=0)$
- When $(x=2\pi): (t=0)$
Because $(\sin x)$ rises from $(0\to 1\to 0)$ on $([0,\pi])$ and then goes $(0\to -1\to 0)$ on $([\pi,2\pi])$, the substitution must be split:
$$
I=\int_{0}^{\pi}(1-\sin^{2}x)^{2}\cos x\;dx+\int_{\pi}^{2\pi}(1-\sin^{2}x)^{2}\cos x\;dx.
$$
Apply the substitution separately:
- For $(0\to \pi)$:โ$(t:0\to 0)$ but the path goes up to $(1)$ and back, so it becomes
$$
\int_{0}^{1}(1-t^{2})^{2}\;dt+\int_{1}^{0}(1-t^{2})^{2}\;dt
$$
which cancel each other. - For $(\pi\to2\pi)$:โ$(t:0\to -1\to 0)$, giving
$$
\int_{0}^{-1}(1-t^{2})^{2}\;dt+\int_{-1}^{0}(1-t^{2})^{2}\;dt,
$$
which again cancel.
So the total integral is
$$
I=0.
$$
Final Answer
$$
\boxed{0}
$$
Reason: $(\cos^{5}x)$ is an odd function of $(\sin x)$ over a full $(0)$ to $(2\pi)$ cycle, giving net area $(0)$.
NCERT Question 15: Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\cos x-\sin x}{1+\sin x\cos x}\;dx$$
Solution
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\cos x-\sin x}{1+\sin x\cos x}\;dx$$
Use the symmetry property $\displaystyle\int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx$ with $a=\dfrac{\pi}{2}$.
Compute the integrand at $\dfrac{\pi}{2}-x$:
$$
f\left(\frac{\pi}{2}-x\right)
=\frac{\cos\left(\frac{\pi}{2}-x\right)-\sin\left(\frac{\pi}{2}-x\right)}{1+\sin\left(\frac{\pi}{2}-x\right)\cos\left(\frac{\pi}{2}-x\right)}$$
$$f\left(\frac{\pi}{2}-x\right)=\frac{\sin x-\cos x}{1+\sin x\cos x}=-\frac{\cos x-\sin x}{1+\sin x\cos x}$$
$$f\left(\frac{\pi}{2}-x\right)=-f(x).$$
Thus $f\left(\dfrac{\pi}{2}-x\right)=-f(x)$, so the integrand is odd on the symmetric interval $[0\;\dfrac{\pi}{2}]$.
When an integrand satisfies $f(a-x)=-f(x)$ on $[0\;a]$ the definite integral over $[0\;a]$ is $0$. Therefore
$$I=0.$$
Final Result
$$\boxed{\;I=0\;}$$
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NCERT Question 16: Prove that
$$
I=\int_{0}^{\dfrac{\pi}{2}}\ln(\sin x)dx =-\frac{\pi}{2}\ln2\
$$
Solution :
Let
$$
I=\int_{0}^{\dfrac{\pi}{2}}\ln(\sin x)dx.
$$
Using the identity $(\sin x\cos x=\tfrac12\sin2x)$ we get
$$
\ln(\sin x)+\ln(\cos x)=\ln\Big(\frac{1}{2}\Big)+\ln(\sin2x).
$$
Integrate both sides from $(0)$ to $(\dfrac{\pi}{2})$. The left side gives
$$
\int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx+\int_{0}^{\frac{\pi}{2}}\ln(\cos x)dx=I+I=2I.
$$
The right side gives
$$
\int_{0}^{\frac{\pi}{2}}\ln\Big(\tfrac{1}{2}\Big)dx+\int_{0}^{\frac{\pi}{2}}\ln(\sin2x)dx
=\frac{\pi}{2}\ln\Big(\tfrac{1}{2}\Big)+\int_{0}^{\frac{\pi}{2}}\ln(\sin2x)dx.
$$
Evaluate the last integral by the substitution $(t=2x)$ :
$$
\int_{0}^{\frac{\pi}{2}}\ln(\sin2x)dx=\frac{1}{2}\int_{0}^{\pi}\ln(\sin t)dt
=\frac{1}{2}\cdot 2\int_{0}^{\frac{\pi}{2}}\ln(\sin t)dt=\frac{1}{2}\cdot 2I=I.
$$
So the right side equals $(\dfrac{\pi}{2}\ln\big(\tfrac{1}{2}\big)+I)$. Therefore
$$ 2I=\frac{\pi}{2}\ln\Big(\tfrac{1}{2}\Big)+I. $$
$$I=\frac{\pi}{2}\ln\Big(\tfrac{1}{2}\Big)= -\frac{\pi}{2}\ln 2$$
Final Result
$$\boxed{\; \displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\sin x)dx=-\frac{\pi}{2}\ln 2\; }$$
Similarly :
$$\boxed{\; \displaystyle \int_{0}^{\frac{\pi}{2}}\ln(\cos x)dx=-\frac{\pi}{2}\ln 2\;}$$
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NCERT Question No.16 : Evaluate the integral
$$\int_{0}^{\pi}\log(1+\cos x)\;dx$$
Solution
$$\int_{0}^{\pi}\log(1+\cos x)\;dx$$
Use the identity
$$1+\cos x=2\cos^2\frac{x}{2}.$$
So,
$$
\log(1+\cos x)
=\log\left(2\cos^2\frac{x}{2}\right)
=\log 2+2\log\left(\cos\frac{x}{2}\right).
$$
Thus the integral becomes
$$
I=\int_{0}^{\pi}\left(\log 2+2\log\left(\cos\frac{x}{2}\right)\right)dx.
$$
Split it:
$$
I=\log 2\int_{0}^{\pi}dx+2\int_{0}^{\pi}\log\left(\cos\frac{x}{2}\right)dx.
$$
Compute each part:
$$
\int_{0}^{\pi}dx=\pi.
$$
- Substitute $(x=2t)\;$ so $(dx=2dt)\;$ limits $(x=0\to t=0)\; (x=\pi\to t=\frac{\pi}{2})$ :
$$
\int_0^\pi \log\left(\cos\frac{x}{2}\right)dx
=\int_0^{\frac{\pi}{2}}2\log(\cos t)\;dt
=2\int_0^{\frac{\pi}{2}}\log(\cos t)\;dt.
$$
Use the standard result (already prove in above question) :
$$
\int_0^{\frac{\pi}{2}}\log(\cos t)\;dt=-\frac{\pi}{2}\log 2.
$$
Thus,
$$
\int_0^\pi \log\left(\cos\frac{x}{2}\right)dx
=2\left(-\frac{\pi}{2}\log 2\right)
=-\pi\log 2.
$$
Now substitute back:
$$
I=\pi\log 2+2(-\pi\log 2)
=-\pi\log 2.
$$
Final Result
$$
\boxed{\;I=-\pi\log 2\;}
$$
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NCERT Question No.17 : Evaluate the integral
$$I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\;dx.$$
Solution :
$$I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\;dx.$$
Step 1: Write the integral
$$I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\;dx \tag{1}$$
Step 2: Apply the property
Using
$$\int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx$$
replace $x$ by $(a-x)$:
$$I=\int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\;dx \tag{2}$$
Step 3: Add (1) and (2)
$$
2I=\int_{0}^{a}\left(
\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}
+
\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}
\right)\;dx.
$$
Simplify the integrand:
$$
\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}=1.
$$
So,
$$2I=\int_{0}^{a}1\;dx=a.$$
Step 4: Final Result
$$\boxed{\;I=\dfrac{a}{2}\;}$$
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NCERT Question No.18 : Evaluate the integral
$$I=\int_{0}^{4}\lvert x-1\rvert\;dx.$$
Solution :
$$I=\int_{0}^{4}\lvert x-1\rvert\;dx.$$
Step 1: Remove the modulus
The expression inside the modulus is
- $ | x -1 | = -(x -1) = 1 – x : 0\le x<1$
- $| x -1 | = (x -1) : 1\le x\le 4$
So split the integral at $x=1$:
$$
I=\int_{0}^{1}(1-x)\;dx+\int_{1}^{4}(x-1)\;dx.
$$
Step 2: Evaluate each part
First integral
$$
\int_{0}^{1}(1-x)\;dx
=\left[x-\frac{x^{2}}{2}\right]_{0}^{1}
=1-\frac12=\frac12.
$$
Second integral
$$
\int_{1}^{4}(x-1)\;dx
=\left[\frac{x^{2}}{2}-x\right]_{1}^{4}
=\left(8-4\right)-\left(\frac12-1\right)
=4+\frac12
=\frac{9}{2}.
$$
Step 3: Add the results
$$
I=\frac12+\frac{9}{2}=5.
$$
Final Result
$$\boxed{5}$$
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NCERT Question 19: Show that
$$\int_{0}^{a} f(x)\;g(x)\;dx=2\int_{0}^{a} f(x)\;dx$$
if $(f(x)=f(a-x))$ for all $(x\in[0\;a])$ and $(g(x)+g(a-x)=4)$ for all $(x\in[0\;a])$.
Solution
Let
$$I=\int_{0}^{a} f(x)\;g(x)\;dx.$$
Replace $(x)$ by $(a-x)$ and use $(dx\to -dx)$ (or perform the standard change of variable $(u=a-x))$ to get a second expression for (I):
$$
I=\int_{0}^{a} f(a-x)\;g(a-x)\;dx.
$$
Because $(f(a-x)=f(x))$ by hypothesis, this becomes
$$
I=\int_{0}^{a} f(x)\;g(a-x)\;dx.
$$
Add the two representations of (I):
$$
2I=\int_{0}^{a} f(x)\bigl(g(x)+g(a-x)\bigr)\;dx.
$$
Use the given relation $(g(x)+g(a-x)=4)$:
$$
2I=\int_{0}^{a} f(x)\cdot 4\;dx=4\int_{0}^{a} f(x)\;dx.
$$
Divide both sides by (2):
$$
I=2\int_{0}^{a} f(x)\;dx.
$$
Final Result
$$\boxed{\;\displaystyle \int_{0}^{a} f(x)\;g(x)\;dx=2\int_{0}^{a} f(x)\;dx\;}$$
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NCERT Question No.20 : The value of integral is
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\Bigl(x^{3}+x\cos x+\tan^{5}x+1\Bigr)\;dx$$
$ (A) 0 \qquad (B) 2 \qquad (C) ฯ \qquad (D) 1 $
Solution
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\Bigl(x^{3}+x\cos x+\tan^{5}x+1\Bigr)\;dx$$
All terms except the constant are odd functions:
- $x^{3}$ is odd
- $x\cos x$ is odd
- $\tan^{5}x$ is odd
For any odd function $f(x)$ on symmetric limits,
$$\int_{-a}^{a}f(x)\;dx=0.$$
Thus,
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^{3}\;dx=0,\qquad
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\cos x\;dx=0,\qquad
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\tan^{5}x\;dx=0.$$
Only the constant term remains:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1\;dx
=\Bigl[x\Bigr]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
=\frac{\pi}{2}-\Bigl(-\frac{\pi}{2}\Bigr)=\pi.$$
Final Result
$$\boxed{\pi}$$
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NCERT Question 21: The value of integral is
$$I=\int_{0}^{\dfrac{\pi}{2}}\ln\left(\frac{4+3\sin x}{4+3\cos x}\right)\;dx$$
$ (A) 0 \qquad (B) 2 \qquad (C) ฯ \qquad (D) 1 $
Solution
$$I=\int_{0}^{\dfrac{\pi}{2}}\ln\left(\frac{4+3\sin x}{4+3\cos x}\right)\;dx$$
Write the integrand as a difference of logs:
$$I=\int_{0}^{\dfrac{\pi}{2}}\ln(4+3\sin x)dx-\int_{0}^{\dfrac{\pi}{2}}\ln(4+3\cos x)dx.$$
Use the substitution $x\to \dfrac{\pi}{2}-x$ in the second integral. Since $\cos x=\sin\big(\dfrac{\pi}{2}-x\big)$ and the substitution preserves the limits $0\to\dfrac{\pi}{2}$, we get
$$\int_{0}^{\dfrac{\pi}{2}}\ln(4+3\cos x)dx=\int_{0}^{\dfrac{\pi}{2}}\ln!\big(4+3\sin x\big)dx.$$
Therefore the two integrals are equal and cancel:
Final Result
$$\boxed{I=0}$$
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