Anand Classes provides detailed NCERT Solutions for Integrals Exercise 7.10 of Chapter 7 for Class 12 Mathematics, offering step-by-step explanations to help students understand and apply advanced integration techniques effectively. These Set-1 notes are based on the latest NCERT syllabus and are ideal for revision, doubt-clearing, and board exam preparation. Click the print button to download study material and notes.
NCERT Question 1: Evaluate the integral
$$\int_{0}^{\dfrac{\pi}{2}} \cos^{2}x\;dx$$
Solution
Let the given integral be
$$I=\int_{0}^{\dfrac{\pi}{2}} \cos^{2}x\;dx \quad ……..(1)$$
Using the property $\int_{0}^{a} f(x)\;dx=\int_{0}^{a} f(a-x)\;dx$, we get
$$I=\int_{0}^{\dfrac{\pi}{2}} \cos^{2}\left(\frac{\pi}{2}-x\right)\;dx$$
Since $\cos\left(\dfrac{\pi}{2}-x\right)=\sin x$, this becomes
$$I=\int_{0}^{\dfrac{\pi}{2}} \sin^{2}x\;dx \quad ………(2)$$
Adding (1) and (2):
$$I+I=\int_{0}^{\dfrac{\pi}{2}} \cos^{2}x\;dx + \int_{0}^{\dfrac{\pi}{2}} \sin^{2}x\;dx$$
$$2I=\int_{0}^{\dfrac{\pi}{2}} \left(\sin^{2}x+\cos^{2}x\right)\;dx$$
Using the identity $\sin^{2}x+\cos^{2}x = 1$
$$2I=\int_{0}^{\dfrac{\pi}{2}} 1\;dx$$
$$2I=\left[x\right]_{0}^{\dfrac{\pi}{2}}$$
$$2I=\dfrac{\pi}{2}$$
$$I=\dfrac{\pi}{4}$$
Final Result
$$\boxed{\dfrac{\pi}{4}}$$
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NCERT Question 2: Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\;dx$$
Solution
Let $$I=\displaystyle\int_{0}^{\dfrac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\;dx.$$
Use the symmetry property $\displaystyle\int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx$ with $a=\dfrac{\pi}{2}$.
Applying it gives
$$I=\int_{0}^{\dfrac{\pi}{2}}\dfrac{\sqrt{\sin\left(\dfrac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\dfrac{\pi}{2}-x\right)}+\sqrt{\cos\left(\dfrac{\pi}{2}-x\right)}}\;dx
=\int_{0}^{\dfrac{\pi}{2}}\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\;dx.$$
Add the two forms.
Adding the original and the transformed integral:
$$
2I=\int_{0}^{\dfrac{\pi}{2}}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\;dx
+\int_{0}^{\dfrac{\pi}{2}}\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\;dx
=\int_{0}^{\dfrac{\pi}{2}}1\;dx.
$$
Evaluate the right-hand side.
$$
2I=\left[x\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}.
$$
Solve for $I$.
$$
I=\frac{\pi}{4}.
$$
Final Result
$$\boxed{\;I=\frac{\pi}{4}\;}$$
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NCERT Question 3: Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\;dx$$
Solution
Let
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\;dx\qquad…………..(1)$$
Use the symmetry property $\displaystyle\int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx$ with $a=\dfrac{\pi}{2}$. Applying it gives
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin^{3/2}\big(\dfrac{\pi}{2}-x\big)}{\sin^{3/2}\big(\dfrac{\pi}{2}-x\big)+\cos^{3/2}\big(\dfrac{\pi}{2}-x\big)}\;dx$$
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\cos^{3/2}x}{\cos^{3/2}x+\sin^{3/2}x}\;dx\qquad………(2)$$
Add (1) and (2).
$$
2I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\;dx
+\int_{0}^{\dfrac{\pi}{2}}\frac{\cos^{3/2}x}{\cos^{3/2}x+\sin^{3/2}x}\;dx $$
$$ I=\int_{0}^{\dfrac{\pi}{2}}1\;dx. $$
Hence
$$
2I=\left[x\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}\quad\Rightarrow\quad I=\frac{\pi}{4}.
$$
Final Result
$$\boxed{\;I=\frac{\pi}{4}\;}$$
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NCERT Question 4 : Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin^{5}x}{\sin^{5}x+\cos^{5}x}\;dx$$
Solution
Let
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin^{5}x}{\sin^{5}x+\cos^{5}x}\;dx$$
Use the symmetry property
$ \displaystyle\int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx$ with $a=\dfrac{\pi}{2}$.
Apply it to $I$ by replacing $x$ with $\dfrac{\pi}{2}-x$. Since $\sin\big(\dfrac{\pi}{2}-x\big)=\cos x$ and $\cos\big(\dfrac{\pi}{2}-x\big)=\sin x$, we get
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\cos^{5}x}{\cos^{5}x+\sin^{5}x}\;dx.$$
Add the two expressions for $I$.
$$
2I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin^{5}x}{\sin^{5}x+\cos^{5}x}\;dx
+\int_{0}^{\dfrac{\pi}{2}}\frac{\cos^{5}x}{\cos^{5}x+\sin^{5}x}\;dx
=\int_{0}^{\dfrac{\pi}{2}}1\;dx.
$$
Evaluate the right-hand side.
$$
2I=\left[x\right]_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}.
$$
Solve for $I$.
Final Result
$$\boxed{\;I=\frac{\pi}{4}\;}$$
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NCERT Question 5 : Evaluate the integral
$$\int_{-5}^{5} \lvert x+2 \rvert \; dx$$
Solution
Let
$$I=\int_{-5}^{5} \lvert x+2 \rvert \; dx \quad \text{…(i)}$$
To remove the modulus, first find where the expression inside the modulus is zero:
$$x+2=0 \;\Rightarrow\; x=-2.$$
Since $-2 \in (-5,5)$, we split the integral at $x=-2$.
From (i),
$$I=\int_{-5}^{5} \lvert x+2 \rvert \; dx$$
$$I=\int_{-5}^{5} \lvert x+2 \rvert \; dx= \int_{-5}^{-2} \lvert x+2 \rvert \; dx + \int_{-2}^{5} \lvert x+2 \rvert \; dx.$$
Now apply the definition of modulus:
- For $x < -2$, $x+2$ is negative, so $\lvert x+2 \rvert = -(x+2)$
- For $x > -2$, $x+2$ is positive, so $\lvert x+2 \rvert = x+2$
Thus the integral becomes:
$$I=\int_{-5}^{-2} -(x+2) \; dx + \int_{-2}^{5} (x+2) \; dx.$$
Evaluate each part
First integral
$$\int_{-5}^{-2} -(x+2) \; dx = \int_{-5}^{-2} (-x-2)\; dx$$
$$\int_{-5}^{-2} -(x+2) \; dx= \left[-\frac{x^{2}}{2}-2x\right]_{-5}^{-2}$$
$$\int_{-5}^{-2} -(x+2) \; dx=\left(-2-(-4)\right)-\left(-\frac{25}{2}+10\right)$$
$$\int_{-5}^{-2} -(x+2) \; dx=2 – \left(-\frac{5}{2}\right)=\frac{9}{2}.$$
Second integral
$$\int_{-2}^{5} (x+2)\; dx=\left[\frac{x^{2}}{2}+2x\right]_{-2}^{5}$$
$$\int_{-2}^{5} (x+2)\; dx=\left(\frac{25}{2}+10\right)-\left(2-4\right)$$
$$\int_{-2}^{5} (x+2)\; dx=\frac{45}{2}+2=\frac{49}{2}.$$
Total value of Integration
$$I=\int_{-5}^{-2} -(x+2) \; dx + \int_{-2}^{5} (x+2) \; dx=\frac{9}{2}+\frac{49}{2}=\frac{58}{2}=29.$$
$$I=\int_{-5}^{5} \lvert x+2 \rvert \; dx = 29$$
Final Answer
$$\boxed{29}$$
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NCERT Question 6: Evaluate the integral
$$\int_{2}^{8} \lvert x-5\rvert \; dx$$
Solution
Let
$$I=\int_{2}^{8} \lvert x-5\rvert \; dx.$$
To remove the modulus, find where the expression inside it becomes zero:
$$x-5=0 \quad \Rightarrow \quad x=5.$$
Since (5) lies between the limits (2) and (8), split the integral:
$$I=\int_{2}^{5} \lvert x-5\rvert \; dx + \int_{5}^{8} \lvert x-5\rvert \; dx.$$
For $(x<5), (x-5)$ is negative, so $(\lvert x-5\rvert = -(x-5) = (5-x)$.
For $(x>5), (x-5)$ is positive, so $(\lvert x-5\rvert = (x-5)$.
Thus,
$$I=\int_{2}^{5} (5-x)\; dx + \int_{5}^{8} (x-5)\; dx.$$
Compute each part
1. First integral
$$\int_{2}^{5} (5-x)\; dx = \left[5x – \frac{x^{2}}{2}\right]_{2}^{5}$$
Evaluate:
At $(x=5) : (25 – \frac{25}{2})$
At $(x=2): (10 – 2)$
Hence
$$\int_{2}^{5} (5-x)\; dx = \left[5x – \frac{x^{2}}{2}\right]_{2}^{5}=\left(25 – \frac{25}{2}\right) – (10 – 2)$$
$$\int_{2}^{5} (5-x)\; dx= \left(\frac{25}{2}\right) – 8= \frac{25 – 16}{2}= \frac{9}{2}.$$
$$\int_{2}^{5} (5-x)\; dx =\frac{9}{2} $$
2. Second integral
$$\int_{5}^{8} (x-5)\; dx
= \left[\frac{x^{2}}{2} – 5x\right]_{5}^{8}$$
Evaluate:
At $(x=8): (\frac{64}{2} – 40 = 32 – 40 = -8)$
At $(x=5): (\frac{25}{2} – 25 = -\frac{25}{2})$
Difference:
$$\int_{5}^{8} (x-5)\; dx
= \left[\frac{x^{2}}{2} – 5x\right]_{5}^{8}=-8 – \left(-\frac{25}{2}\right)
= -8 + \frac{25}{2}
= \frac{25 – 16}{2}
= \frac{9}{2}.$$
Add both parts
$$I=\int_{2}^{5} (5-x)\; dx + \int_{5}^{8} (x-5)\; dx.=\frac{9}{2} + \frac{9}{2} = 9$$
Final Result
$$\boxed{9}$$
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NCERT Question 7: Evaluate the integral
$$I=\int_{0}^{1} x(1-x)^{n}\;dx$$
Solution
Let
$$I=\int_{0}^{1}x(1-x)^{n}\;dx $$
Use the definite–integral property
$$\int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx$$
Here, $a=1$ and $f(x)=x(1-x)^{n}$.
Applying the property:
$$I=\int_{0}^{1}(1-x)\big(1-(1-x)\big)^{n}\;dx$$
Simplify inside the integrand:
Since $1-(1-x)=x$,
$$I=\int_{0}^{1}(1-x)x^{n}\;dx $$
$$I=\int_{0}^{1}(x^{n}-x^{n+1})\;dx $$
$$I=\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_{0}^{1}$$
Thus:
$$
I=\left(\frac{1}{n+1}-\frac{1}{n+2}\right)
$$
$$I=\frac{1}{(n+1)(n+2)}$$
Final Result
$$\boxed{I=\frac{1}{(n+1)(n+2)}}$$
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NCERT Question 8: Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{4}}\ln\bigl(1+\tan x\bigr)\;dx$$
Solution
Let
$$I=\int_{0}^{\dfrac{\pi}{4}}\ln\bigl(1+\tan x\bigr)\;dx$$
Use the symmetry property $$\displaystyle\int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx) $$ with $ a=\dfrac{\pi}{4}$
Apply $(x\mapsto \dfrac{\pi}{4}-x)$.
Note
$$
\tan\Big(\frac{\pi}{4}-x\Big)=\frac{1-\tan x}{1+\tan x}.
$$
Hence
$$
1+\tan\Big(\frac{\pi}{4}-x\Big)=\frac{2}{1+\tan x}.
$$
Therefore
$$
I=\int_{0}^{\dfrac{\pi}{4}}\ln\Big(1+\tan\Big(\frac{\pi}{4}-x\Big)\Big)\;dx
=\int_{0}^{\dfrac{\pi}{4}}\ln\Big(\frac{2}{1+\tan x}\Big)\;dx.
$$
Split the log:
$$
I=\int_{0}^{\dfrac{\pi}{4}}\bigl(\ln 2-\ln(1+\tan x)\bigr)\;dx
=\frac{\pi}{4}\ln 2 – I.
$$
Solve for (I).
$$
2I=\frac{\pi}{4}\ln 2\quad\Longrightarrow\quad I=\frac{\pi}{8}\ln 2.
$$
Final Result
$$\boxed{\;I=\frac{\pi}{8}\ln 2\;}$$
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NCERT Question 9: Evaluate the integral
$$\int_{0}^{2} x\sqrt{2-x}\; dx$$
Solution :
$$\int_{0}^{2} x\sqrt{2-x}\; dx$$
Use the property
$$\int_0^a f(x)\; dx = \int_0^a f(a-x)\; dx.$$
Let
$$I = \int_{0}^{2} x\sqrt{2-x}\; dx.$$
Apply the property by replacing $(x)$ with $(2-x)$:
$$
I = \int_{0}^{2} (2-x)\sqrt{x}\; dx.
$$
$$I = \int_{0}^{2} (2x^{1/2}-x^{3/2})\; dx.$$
$$
I = \left[\frac{4}{3}x^{3/2} – \frac{2}{5}x^{5/2}\right]_{0}^{2}
$$
$$
I = \frac{4}{3}(2\sqrt{2}) – \frac{2}{5}(4\sqrt{2})
$$
$$
I = \frac{8\sqrt{2}}{3} – \frac{8\sqrt{2}}{5}
$$
$$
I = 8\sqrt{2}\left(\frac{1}{3} – \frac{1}{5}\right)
$$
$$
I = 8\sqrt{2}\left(\frac{2}{15}\right)
$$
$$
\boxed{I = \frac{16\sqrt{2}}{15}}
$$
NCERT Question 10: Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{2}}\bigl(2\ln\sin x-\ln\sin 2x\bigr)\;dx$$
$$I=\int_{0}^{\dfrac{\pi}{2}}\bigl(2\ln\sin x-\ln\sin 2x\bigr)\;dx$$
Solution :
Step 1 — simplify the integrand.
Use $\sin2x=2\sin x\cos x$ so
$$\ln\sin2x=\ln2+\ln\sin x+\ln\cos x.$$
Hence
$$2\ln\sin x-\ln\sin2x=\ln\sin x-\ln\cos x-\ln2=\ln(\tan x)-\ln2.$$
So
$$I=\int_{0}^{\dfrac{\pi}{2}}\ln(\tan x)\;dx-\int_{0}^{\dfrac{\pi}{2}}\ln2\;dx.$$
Step 2 — evaluate the $\ln\tan x$ integral by symmetry.
Let
$$J=\int_{0}^{\dfrac{\pi}{2}}\ln(\tan x)\;dx.$$
Apply the substitution $x\mapsto \dfrac{\pi}{2}-x$; then $\tan\big(\dfrac{\pi}{2}-x\big)=\cot x$ and
$$J=\int_{0}^{\dfrac{\pi}{2}}\ln(\cot x)\;dx.$$
Add the two expressions:
$$2J=\int_{0}^{\dfrac{\pi}{2}}\bigl(\ln\tan x+\ln\cot x\bigr)\;dx=\int_{0}^{\dfrac{\pi}{2}}\ln 1\;dx=0\;$$
so
$$J=0$$
Step 3 — finish. The remaining term is
$$I=0-\int_{0}^{\dfrac{\pi}{2}}\ln2\;dx=-\frac{\pi}{2}\ln2.$$
Final Result
$$\boxed{\;I=-\frac{\pi}{2}\ln 2\;}$$
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