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NCERT Question 8: Evaluate the Integral
$$\displaystyle \int \frac{1 – \cos(x)}{1 + \cos(x)}\; dx$$
Solution
$$\displaystyle \int \frac{1 – \cos(x)}{1 + \cos(x)}\; dx$$
First, simplify the integrand using trigonometric identities.
We know:
$$1 – \cos(x) = 2\sin^2\left(\frac{x}{2}\right)$$
$$1 + \cos(x) = 2\cos^2\left(\frac{x}{2}\right)$$
Substitute these values:
$$\frac{1 – \cos(x)}{1 + \cos(x)} = \frac{2\sin^2\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)} = \tan^2\left(\frac{x}{2}\right)$$
Thus,
$$\int \frac{1 – \cos(x)}{1 + \cos(x)}\; dx = \int \tan^2\left(\frac{x}{2}\right)\; dx$$
We know the identity:
$$\tan^2\theta = \sec^2\theta – 1$$
So,
$$\int \tan^2\left(\frac{x}{2}\right)\; dx = \int \left[\sec^2\left(\frac{x}{2}\right) – 1\right]\; dx$$
Now integrate term by term:
Let $u = \dfrac{x}{2}$, hence $du = \dfrac{1}{2}\; dx$ or $dx = 2\; du$.
Substitute:
$$\int \left[\sec^2(u) – 1\right]2\; du = 2\int \sec^2(u)\; du – 2\int 1\; du$$
Integrate:
$$= 2\tan(u) – 2u + C$$
Substitute back $u = \dfrac{x}{2}$:
$$= 2\tan\left(\frac{x}{2}\right) – x + C$$
Final Answer
$$\boxed{\displaystyle \int \frac{1 – \cos(x)}{1 + \cos(x)}\; dx = 2\tan\left(\frac{x}{2}\right) – x + C}$$
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NCERT Question 9: Evaluate the Integral
$$\displaystyle \int \frac{\cos(x)}{1+\cos(x)}\; dx$$
Solution
$$\displaystyle \int \frac{\cos(x)}{1+\cos(x)}\; dx$$
Rewrite the integrand by splitting the fraction:
$$\frac{\cos(x)}{1+\cos(x)}=\frac{1+\cos(x)-1}{1+\cos(x)}=1-\frac{1}{1+\cos(x)}$$
So
$$\int \frac{\cos(x)}{1+\cos(x)}\; dx=\int 1\; dx-\int \frac{1}{1+\cos(x)}\; dx = x-\int \frac{1}{1+\cos(x)}\; dx + C$$
Use the half-angle identity $1+\cos(x)=2\cos^2\bigl(\dfrac{x}{2}\bigr)$, hence
$$\frac{1}{1+\cos(x)}=\frac{1}{2\cos^2\bigl(\dfrac{x}{2}\bigr)}=\dfrac{1}{2}\sec^2\bigl(\dfrac{x}{2}\bigr)$$
Thus
$$\int \frac{1}{1+\cos(x)}\; dx=\dfrac{1}{2}\int \sec^2\bigl(\dfrac{x}{2}\bigr)\; dx$$
Let $u=\dfrac{x}{2}\Rightarrow du=\dfrac{1}{2}\;dx\Rightarrow dx=2\;du$. Then
$$\dfrac{1}{2}\int \sec^2\bigl(\dfrac{x}{2}\bigr)\; dx
=\dfrac{1}{2}\cdot 2\int \sec^2(u)\; du
=\int \sec^2(u)\; du
=\tan(u)$$
Substitute back $u=\dfrac{x}{2}$ to get $\tan\bigl(\dfrac{x}{2}\bigr)$.
Therefore the integral is
$$\boxed{\displaystyle \int \frac{\cos(x)}{1+\cos(x)}\; dx = x-\tan\bigl(\dfrac{x}{2}\bigr)+C}$$
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NCERT Question 10: Evaluate the Integral
$$\displaystyle \int \sin^4(x)\; dx$$
Solution
$$\displaystyle \int \sin^4(x)\; dx$$
Use the power-reduction identity:
$$\sin^2(x) = \frac{1 – \cos(2x)}{2}$$
Hence,
$$\sin^4(x) = \left(\frac{1 – \cos(2x)}{2}\right)^2 = \frac{1}{4}(1 – 2\cos(2x) + \cos^2(2x))$$
Now substitute $\cos^2(2x) = \dfrac{1 + \cos(4x)}{2}$:
$$\sin^4(x) = \frac{1}{4}\left[1 – 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right]$$
Simplify:
$$\sin^4(x) = \frac{1}{4}\left(\frac{3}{2} – 2\cos(2x) + \frac{1}{2}\cos(4x)\right)$$
$$\sin^4(x) = \frac{3}{8} – \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)$$
Now integrate each term:
$$\int \sin^4(x)\; dx = \int \left(\frac{3}{8} – \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\right)\; dx$$
Integrate term by term:
$$= \frac{3x}{8} – \frac{1}{2} \times \frac{\sin(2x)}{2} + \frac{1}{8} \times \frac{\sin(4x)}{4} + C$$
Simplify:
$$\boxed{\displaystyle \int \sin^4(x)\; dx = \frac{3x}{8} – \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C}$$
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NCERT Question 11: Evaluate the Integral
$$\displaystyle \int \cos^4(2x)\; dx$$
Solution
$$\displaystyle \int \cos^4(2x)\; dx$$
Use the power-reduction identity:
$$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$
So,
$$\cos^4(2x) = \left(\frac{1 + \cos(4x)}{2}\right)^2 = \frac{1}{4}\left(1 + 2\cos(4x) + \cos^2(4x)\right)$$
Now, apply the identity again to $\cos^2(4x)$:
$$\cos^2(4x) = \frac{1 + \cos(8x)}{2}$$
Substitute it:
$$\cos^4(2x) = \frac{1}{4}\left[1 + 2\cos(4x) + \frac{1 + \cos(8x)}{2}\right]$$
Simplify:
$$\cos^4(2x) = \frac{1}{4}\left(\frac{3}{2} + 2\cos(4x) + \frac{1}{2}\cos(8x)\right)$$
$$\cos^4(2x) = \frac{3}{8} + \frac{1}{2}\cos(4x) + \frac{1}{8}\cos(8x)$$
Now integrate each term:
$$\int \cos^4(2x)\; dx = \int \left(\frac{3}{8} + \frac{1}{2}\cos(4x) + \frac{1}{8}\cos(8x)\right)\; dx$$
Integrate term by term:
$$= \frac{3x}{8} + \frac{1}{2} \times \frac{\sin(4x)}{4} + \frac{1}{8} \times \frac{\sin(8x)}{8} + C$$
Simplify:
$$\boxed{\displaystyle \int \cos^4(2x)\; dx = \frac{3x}{8} + \frac{1}{8}\sin(4x) + \frac{1}{64}\sin(8x) + C}$$
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NCERT Question 12: Evaluate the Integral
$$\displaystyle \int \frac{\sin^2(x)}{1+\cos(x)}\; dx$$
Solution
$$\displaystyle \int \frac{\sin^2(x)}{1+\cos(x)}\; dx$$
Simplify the integrand using trigonometric identities.
We know:
$$\sin^2(x) = 1 – \cos^2(x)$$
So,
$$\frac{\sin^2(x)}{1+\cos(x)} = \frac{1 – \cos^2(x)}{1+\cos(x)}$$
Factorize the numerator:
$$1 – \cos^2(x) = (1 – \cos(x))(1 + \cos(x))$$
Hence,
$$\frac{\sin^2(x)}{1+\cos(x)} = 1 – \cos(x)$$
Now the integral becomes:
$$\int \frac{\sin^2(x)}{1+\cos(x)}\; dx = \int (1 – \cos(x))\; dx$$
Integrate term by term:
$$= \int 1\; dx – \int \cos(x)\; dx$$
$$= x – \sin(x) + C$$
Final Answer
$$\boxed{\displaystyle \int \frac{\sin^2(x)}{1+\cos(x)}\; dx = x – \sin(x) + C}$$
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NCERT Question 13: Evaluate the Integral
$$\displaystyle \int \frac{\cos(2x)-\cos(2\alpha)}{\cos(x)-\cos(\alpha)}\; dx$$
Solution
$$\displaystyle \int \frac{\cos(2x)-\cos(2\alpha)}{\cos(x)-\cos(\alpha)}\; dx$$
Use the double-angle identity $\cos(2t)=2\cos^2 t-1$. Then the numerator becomes
$$\cos(2x)-\cos(2\alpha)=2\cos^2 x-1- (2\cos^2\alpha-1)
=2\bigl(\cos^2 x-\cos^2\alpha\bigr).$$
Factor the difference of squares:
$$2\bigl(\cos^2 x-\cos^2\alpha\bigr)=2\bigl(\cos x-\cos\alpha\bigr)\bigl(\cos x+\cos\alpha\bigr).$$
Therefore the integrand simplifies to
$$\frac{\cos(2x)-\cos(2\alpha)}{\cos(x)-\cos(\alpha)}
=2\bigl(\cos x+\cos\alpha\bigr).$$
So the integral becomes
$$\int 2\bigl(\cos x+\cos\alpha\bigr)\; dx
=2\int \cos x\; dx + 2\cos\alpha\int 1\; dx.$$
Integrate termwise:
$$2\int \cos x\; dx = 2\sin x,\qquad 2\cos\alpha\int 1\; dx = 2\cos\alpha\; x.$$
Final Answer
$$\boxed{\displaystyle \int \frac{\cos(2x)-\cos(2\alpha)}{\cos(x)-\cos(\alpha)}\; dx
=2\sin x + 2x\cos\alpha + C}$$
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NCERT Question 14: Evaluate the Integral
$$\displaystyle \int \frac{\cos(x)-\sin(x)}{1+\sin(2x)}\; dx$$
Solution
$$\displaystyle \int \frac{\cos(x)-\sin(x)}{1+\sin(2x)}\; dx$$
Note that
$$1+\sin(2x)=1+2\sin(x)\cos(x)=(\sin(x)+\cos(x))^2.$$
Let
$$u=\sin(x)+\cos(x)\quad\Rightarrow\quad du=(\cos(x)-\sin(x))\; dx.$$
Hence the integral becomes
$$\int \frac{du}{u^2} = -\frac{1}{u}+C.$$
Substituting back,
$$\boxed{\displaystyle \int \frac{\cos(x)-\sin(x)}{1+\sin(2x)}\; dx = -\frac{1}{\sin(x)+\cos(x)} + C}$$
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NCERT Question 15: Evaluate the Integral
$$\displaystyle \int \tan^3(2x)\sec(2x)\; dx$$
Solution
$$\displaystyle \int \tan^3(2x)\sec(2x)\; dx$$
Let
$$I = \int \tan^3(2x)\sec(2x)\; dx$$
We know that $\tan^2(2x) = \sec^2(2x) – 1$.
Hence,
$$I = \int \tan(2x)\bigl(\sec^2(2x) – 1\bigr)\sec(2x)\; dx$$
Simplify:
$$I = \int \tan(2x)\sec^3(2x)\; dx \;- \int \tan(2x)\sec(2x)\; dx$$
Now let $u = \tan(2x)$, then
$$du = 2\sec^2(2x)\; dx \quad \Rightarrow \quad dx = \frac{du}{2\sec^2(2x)}.$$
Substitute into the integral term by term.
First Term:
$$\int \tan(2x)\sec^3(2x)\; dx = \frac{1}{2}\int u\sec(2x)\; du$$
But $\sec^2(2x) = 1 + u^2 \;\Rightarrow\; \sec(2x) = \sqrt{1 + u^2}$.
So,
$$\frac{1}{2}\int u\sqrt{1+u^2}\; du$$
Let $v = 1 + u^2$, then $dv = 2u\; du \Rightarrow u\; du = \dfrac{dv}{2}.$
Thus,
$$\frac{1}{2}\int u\sqrt{1+u^2}\; du = \frac{1}{4}\int v^{1/2}\; dv = \frac{1}{4} \times \frac{2}{3}v^{3/2} = \frac{1}{6}(1 + u^2)^{3/2}$$
Substitute $u = \tan(2x)$:
$$\frac{1}{6}(1 + u^2)^{3/2}=\frac{1}{6}(1 + \tan^2(2x))^{3/2}=\frac{1}{6}(\sec^2(2x))^{3/2}=\frac{1}{6}\sec^3(2x)$$
Hence,
$$\frac{1}{2}\int u\sqrt{1+u^2}\; du =\frac{1}{6}\sec^3(2x)$$
Second Term:
$$\int \tan(2x)\sec(2x)\; dx = \frac{1}{2}\sec(2x) + C$$
Now combine both results:
$$ I=\int \tan(2x)\sec^3(2x)\; dx \;- \int \tan(2x)\sec(2x)\; dx $$
$$ I = \frac{1}{6}\sec^3(2x) – \frac{1}{2}\sec(2x) + C$$
Final Answer
$$\boxed{\displaystyle \int \tan^3(2x)\sec(2x)\; dx = \frac{1}{6}\sec^3(2x) – \frac{1}{2}\sec^2(2x) + C}$$
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