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NCERT Question 31: Evaluate the integral
$$\int \frac{\sin x}{(1 + \cos x)^2}\;dx$$
Solution :
$$\int \frac{\sin x}{(1 + \cos x)^2}\;dx$$
Let
$$1 + \cos x = t$$
Then differentiating,
$$-\sin x\;dx = dt \quad \Rightarrow \quad \sin x\;dx = -dt$$
Substitute in the integral:
$$\int \frac{\sin x}{(1 + \cos x)^2}\;dx = \int \frac{-dt}{t^2} = -\int t^{-2}\;dt$$
Integrate:
$$-\int t^{-2}\;dt = -\left(\frac{t^{-1}}{-1}\right) + C = \frac{1}{t} + C$$
Substitute back $t = 1 + \cos x$:
$$\frac{1}{t} + C = \frac{1}{1 + \cos x} + C$$
Final Answer
$$\boxed{\frac{1}{1 + \cos x} + C}$$
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NCERT Question 32: Evaluate the integral
$$\int \frac{1}{1+\cot x}\;dx$$
Solution :
Rewrite the integrand using $\cot x=\dfrac{\cos x}{\sin x}$
$$\frac{1}{1+\cot x}=\frac{1}{1+\dfrac{\cos x}{\sin x}}=\frac{1}{\dfrac{\sin x+\cos x}{\sin x}}=\frac{\sin x}{\sin x+\cos x}.$$
So
$$\int \frac{1}{1+\cot x}\;dx=\int \frac{\sin x}{\sin x+\cos x}\;dx.$$
Split the integrand by adding and subtracting $\cos x$ in the numerator:
$$\frac{\sin x}{\sin x+\cos x}=\frac{1}{2}\cdot\frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x} $$
$$\frac{\sin x}{\sin x+\cos x}=\frac{1}{2}\Bigg(1+\frac{\sin x-\cos x}{\sin x+\cos x}\Bigg).$$
Thus
$$\int \frac{\sin x}{\sin x+\cos x}\;dx =\frac{1}{2}\int dx+\frac{1}{2}\int\frac{\sin x-\cos x}{\sin x+\cos x}\;dx $$
$$\int \frac{\sin x}{\sin x+\cos x}\;dx =\frac{x}{2}+\frac{1}{2}\int\frac{\sin x-\cos x}{\sin x+\cos x}\;dx.$$
Put
$$t=\sin x+\cos x\quad\Rightarrow\quad dt=(\cos x-\sin x)\;dx=-(\sin x-\cos x)\;dx.$$
Hence $(\sin x-\cos x)\;dx=-dt$ and
$$\frac{1}{2}\int\frac{\sin x-\cos x}{\sin x+\cos x}\;dx
=\frac{1}{2}\int\frac{-dt}{t}=-\frac{1}{2}\ln|t|+C.$$
Substitute back $t=\sin x+\cos x$ to get
$$=\frac{x}{2}-\frac{1}{2}\ln\big|\sin x+\cos x\big|+C.$$
Final Answer
$$\boxed{\;\frac{x}{2}-\frac{1}{2}\ln\big|\sin x+\cos x\big|+C\;}$$
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NCERT Question 33: Evaluate the integral
$$\int \frac{1}{1-\tan x}\;dx$$
Solution :
$$\int \frac{1}{1-\tan x}\;dx$$
We know that $\tan x = \dfrac{\sin x}{\cos x}$, hence
$$\int \frac{1}{1-\tan x}\;dx = \int \frac{1}{1-\dfrac{\sin x}{\cos x}}\;dx = \int \frac{\cos x}{\cos x-\sin x}\;dx.$$
$$\int \frac{\cos x}{\cos x-\sin x}\;dx = \frac{1}{2}\int \frac{2\cos x}{\cos x-\sin x}\;dx $$ $$ = \frac{1}{2}\int \frac{(\cos x-\sin x)+(\sin x+\cos x)}{\cos x-\sin x}\;dx.$$
Separate the terms:
$$I = \frac{1}{2}\int 1\;dx + \frac{1}{2}\int \frac{\sin x+\cos x}{\cos x-\sin x}\;dx.$$
Let $\cos x – \sin x = t \Rightarrow (-\sin x – \cos x)\;dx = dt.$
Then
$$\sin x + \cos x = -\frac{dt}{dx}$$
and the integral becomes
$$ = \frac{1}{2}x + \frac{1}{2}\int \frac{-(dt)}{t} = \frac{1}{2}x – \frac{1}{2}\log|t| + C.$$
Substitute back $t = \cos x – \sin x$:
$$ = \frac{x}{2} – \frac{1}{2}\log|\cos x – \sin x| + C.$$
Hence,
$$\boxed{\;I = \frac{x}{2} – \frac{1}{2}\log|\cos x – \sin x| + C\;}$$
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NCERT Question 34: Evaluate the integral
$$\int \frac{\sqrt{\tan x}}{\sin x\cos x}\;dx$$
Solution :
$$\int \frac{\sqrt{\tan x}}{\sin x\cos x}\;dx$$
Put
$$t=\sqrt{\tan x}\quad\Rightarrow\quad t^2=\tan x.$$
Differentiate both sides:
$$2t\;dt=\sec^2 x\;dx.$$
Solve for $dx$:
$$dx=\frac{2t\;dt}{\sec^2 x}=2t\cos^2 x\;dt.$$
Now substitute into the integrand. Since $\sqrt{\tan x}=t$ we get
$$
\frac{\sqrt{\tan x}}{\sin x\cos x}\;dx
= \frac{t}{\sin x\cos x}\cdot 2t\cos^2 x\;dt
=2t^2\frac{\cos^2 x}{\sin x\cos x}\;dt $$
$$\frac{\sqrt{\tan x}}{\sin x\cos x}\;dx =2t^2\frac{\cos x}{\sin x}\;dt =2t^2\cot x\;dt. $$
But $\tan x=t^2$ so $\cot x=\dfrac{1}{\tan x}=\dfrac{1}{t^2}$. Therefore the integrand becomes simply
$$2t^2\cdot\frac{1}{t^2}\;dt=2\;dt.$$
Thus the integral reduces to
$$\int 2\;dt=2t+C.$$
Substitute back $t=\sqrt{\tan x}$ to obtain the result.
Final Answer
$$\boxed{\;2\sqrt{\tan x}+C\;}$$
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NCERT Question 35: Evaluate the integral
$$\int \frac{(1+\log x)^2}{x}\;dx$$
Solution :
$$\int \frac{(1+\log x)^2}{x}\;dx$$
Let
$$1+\log x=t\quad\Rightarrow\quad \frac{1}{x}\;dx=dt$$
Substituting in the integral,
$$\int \frac{(1+\log x)^2}{x}\;dx=\int t^2\;dt$$
Integrate:
$$\int t^2\;dt=\frac{t^3}{3}+C$$
Now substitute back $t=1+\log x$:
$$I=\frac{(1+\log x)^3}{3}+C$$
Final Answer
$$\boxed{\;\frac{(1+\log x)^3}{3}+C\;}$$
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NCERT Question 36: Evaluate the integral
$$\int \frac{(x+1)(x+\log x)^2}{x}\;dx$$
Solution :
$$\int \frac{(x+1)(x+\log x)^2}{x}\;dx$$
Rewrite the expression:
$$\frac{(x+1)(x+\log x)^2}{x}=\left(1+\frac{1}{x}\right)(x+\log x)^2$$
Let
$$(x+\log x)=t \quad\Rightarrow\quad \left(1+\frac{1}{x}\right)dx=dt$$
Substituting,
$$\int \left(1+\frac{1}{x}\right)(x+\log x)^2\;dx=\int t^2\;dt$$
Integrate:
$$\int t^2\;dt=\frac{t^3}{3}+C$$
Now substitute $t=x+\log x$:
$$=\frac{(x+\log x)^3}{3}+C$$
Final Answer
$$\boxed{\;\frac{(x+\log x)^3}{3}+C\;}$$
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NCERT Question 37: Evaluate the integral
$$\int \frac{x^{3}\;\sin\big(\tan^{-1}x^{4}\big)}{1+x^{8}}\;dx$$
Solution :
$$\int \frac{x^{3}\;\sin\big(\tan^{-1}x^{4}\big)}{1+x^{8}}\;dx$$
Put
$$t = x^{4}\quad\Rightarrow\quad dt = 4x^{3}\:dx\quad\Rightarrow\quad x^{3}\:dx=\frac{dt}{4}.$$
The integral becomes
$$\int \frac{x^{3}\sin(\tan^{-1}x^{4})}{1+x^{8}}\:dx
=\frac{1}{4}\int \frac{\sin(\tan^{-1}t)}{1+t^{2}}\:dt.$$
Now put
$$u=\tan^{-1}t\quad\Rightarrow\quad du=\frac{dt}{1+t^{2}}.$$
Hence the integral reduces to
$$\frac{1}{4}\int \sin u\:du=\frac{1}{4}(-\cos u)+C=-\frac{1}{4}\cos u + C.$$
Substitute back $u=\tan^{-1}t$ and $t=x^{4}$ to get
$$-\frac{1}{4}\cos\big(\tan^{-1}x^{4}\big)+C.$$
We may simplify $\cos(\tan^{-1}x^{4})$: for $u=\tan^{-1}t$ we have $\cos u=\dfrac{1}{\sqrt{1+t^{2}}}$, so with $t=x^{4}$ an equivalent form is
$$-\frac{1}{4}\cdot\frac{1}{\sqrt{1+x^{8}}}+C.$$
Final Answer
$$\boxed{-\frac{1}{4}\cos\big(\tan^{-1}x^{4}\big)+C
=-\frac{1}{4\sqrt{1+x^{8}}}+C}$$
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NCERT Question 38: Evaluate the integral
$$
\int \frac{10x^{9} + 10^{x} \ln 10}{10^{x} + x^{10}}\:dx
$$
Solution
$$
\int \frac{10x^{9} + 10^{x} \ln 10}{10^{x} + x^{10}}\:dx
$$
Let
$$
t = 10^{x} + x^{10}
$$
Then,
$$
dt = (10^{x} \ln 10 + 10x^{9})\;dx
$$
Hence,
$$
\int \frac{10x^{9} + 10^{x} \ln 10}{10^{x} + x^{10}}\;dx = \int \frac{dt}{t}
$$
Therefore,
$$
\int \frac{dt}{t} = \ln |t| + C
$$
Substituting back the value of $t$, we get
$$
\ln |10^{x} + x^{10}| + C
$$
Final Answer : Option (D)
$$
\boxed{\ln(10^{x} + x^{10}) + C\;}
$$
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NCERT Question 39: Evaluate the integral
$$\int \frac{1}{\sin^{2}x\cos^{2}x}\;dx$$
Solution :
$$\int \frac{1}{\sin^{2}x\cos^{2}x}\;dx$$
Use the Pythagorean identity $ \sin^{2}x+\cos^{2}x=1 $ and rewrite the integrand:
$$\frac{1}{\sin^{2}x\cos^{2}x}=\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}$$
$$ \frac{1}{\sin^{2}x\cos^{2}x}=\frac{\sin^{2}x}{\sin^{2}x\cos^{2}x}+\frac{\cos^{2}x}{\sin^{2}x\cos^{2}x}.$$
Simplify each term:
$$\frac{\sin^{2}x}{\sin^{2}x\cos^{2}x}=\frac{1}{\cos^{2}x}=\sec^{2}x, $$
$$ \frac{\cos^{2}x}{\sin^{2}x\cos^{2}x}=\frac{1}{\sin^{2}x}=\csc^{2}x.$$
So the integral becomes
$$\int\frac{1}{\sin^{2}x\cos^{2}x}\;dx=\int\sec^{2}x\;dx+\int\csc^{2}x\;dx.$$
Integrate termwise:
$$\int\sec^{2}x\;dx=\tan x $$
$$ \int\csc^{2}x\;dx=-\cot x.$$
Therefore
$$\int\frac{1}{\sin^{2}x\cos^{2}x}\;dx=\tan x-\cot x + C.$$
Final Answer (choice)
$$\boxed{\;(B)\quad\tan x – \cot x + C\;}$$
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