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NCERT Question 1: Evaluate the Integral
$$\int \sin^2(2x + 5)\; dx$$
Solution:
$$\int \sin^2(2x + 5)\; dx$$
Let $u = 2x + 5$
Then,
$$du = 2\; dx \Rightarrow dx = \frac{du}{2}$$
Substitute in the integral:
$$\int \sin^2(2x + 5)\; dx = \frac{1}{2} \int \sin^2(u)\; du$$
Using the identity $\sin^2(u) = \dfrac{1 – \cos(2u)}{2}$,
$$\frac{1}{2} \int \sin^2(u)\; du = \frac{1}{2} \int \frac{1 – \cos(2u)}{2}\; du$$
$$\frac{1}{2} \int \sin^2(u)\; du = \frac{1}{4} \int (1 – \cos(2u))\; du$$
Integrating term by term,
$$ \frac{1}{4} \int (1 – \cos(2u))\; du= \frac{1}{4} \left( u – \frac{1}{2} \sin(2u) \right) + C$$
Now substitute $u = 2x + 5$,
$$= \frac{1}{4} \left( 2x + 5 – \frac{1}{2} \sin(4x + 10) \right) + C$$
Hence, the required integral is
$$\boxed{\int \sin^2(2x + 5)\; dx = \frac{1}{4} \left( 2x + 5 – \frac{1}{2} \sin(4x + 10) \right) + C}$$
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NCERT Question 2: Evaluate the Integral
$$\int \sin(3x)\cos(4x)\; dx$$
Solution:
$$\int \sin(3x)\cos(4x)\; dx$$
Using the identity $\sin(a)\cos(b) = \frac{1}{2}[\sin(a + b) + \sin(a – b)]$,
$$\int \sin(3x)\cos(4x)\; dx = \frac{1}{2} \int [\sin(7x) + \sin(-x)]\; dx$$
Simplify $\sin(-x) = -\sin(x)$,
$$= \frac{1}{2} \int [\sin(7x) – \sin(x)]\; dx$$
Integrate each term:
$$= \frac{1}{2} \left( -\frac{1}{7} \cos(7x) + \cos(x) \right) + C$$
Simplify,
$$= -\frac{1}{14} \cos(7x) + \frac{1}{2} \cos(x) + C$$
Hence, the required integral is
$$\boxed{\int \sin(3x)\cos(4x)\; dx = -\frac{1}{14}\cos(7x) + \frac{1}{2}\cos(x) + C}$$
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NCERT Question 3: Evaluate the Integral
$$\int \cos(2x)\cos(4x)\cos(6x)\; dx$$
Solution:
$$\int \cos(2x)\cos(4x)\cos(6x)\; dx$$
Use the product-to-sum formula for three cosines:
$$\cos a\cos b\cos c = \frac{1}{4}\bigl(\cos(a+b+c)+\cos(a+b-c)+\\+\cos(a-b+c)+\cos(-a+b+c)\bigr)$$
With $a=2x,\; b=4x,\; c=6x$ we get
$$\cos(2x)\cos(4x)\cos(6x)
= \frac{1}{4}\bigl(\cos(12x)+\cos(0x)+\cos(4x)+\cos(8x)\bigr)$$
$$ \cos(2x)\cos(4x)\cos(6x) = \frac{1}{4}\bigl(\cos(12x)+1+\cos(4x)+\cos(8x)\bigr)$$
Therefore
$$
\int \cos(2x)\cos(4x)\cos(6x)\; dx
= \frac{1}{4}\int\bigl(1+\cos(4x)+\cos(8x)+\cos(12x)\bigr)\; dx $$
$$ = \frac{1}{4}\Bigl(x+\frac{1}{4}\sin(4x)+\frac{1}{8}\sin(8x)+\frac{1}{12}\sin(12x)\Bigr)+C$$
Final Answer
$$\boxed{\; \int \cos(2x)\cos(4x)\cos(6x)\; dx \\= \frac{1}{4}x+\frac{1}{16}\sin(4x)+\frac{1}{32}\sin(8x)+\frac{1}{48}\sin(12x)+C\;}$$
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NCERT Question 4: Evaluate the Integral
$$\displaystyle \int \sin^3(2x + 1)\; dx$$
Solution :
$$\displaystyle \int \sin^3(2x + 1)\; dx$$
Let $u = 2x + 1$
Then,
$$du = 2\; dx \Rightarrow dx = \frac{du}{2}$$
So,
$$\int \sin^3(2x + 1)\; dx = \frac{1}{2} \int \sin^3(u)\; du$$
Now, use the trigonometric identity $\sin^3(u) = \sin(u)\bigl(1 – \cos^2(u)\bigr)$:
$$\frac{1}{2} \int \sin^3(u)\; du = \frac{1}{2} \int \sin(u)\bigl(1 – \cos^2(u)\bigr)\; du$$
Let $\cos(u) = t \Rightarrow -\sin(u)\; du = dt$
Hence,
$$\frac{1}{2} \int \sin(u)\bigl(1 – \cos^2(u)\bigr)\; du = -\frac{1}{2} \int (1 – t^2)\; dt$$
Integrate:
$$-\frac{1}{2} \int (1 – t^2)\; dt =-\frac{1}{2}\left(t – \frac{t^3}{3}\right) + C$$
Substitute back $t = \cos(u)$:
$$-\frac{1}{2}\left(t – \frac{t^3}{3}\right) + C=-\frac{1}{2}\left(\cos(u) – \frac{\cos^3(u)}{3}\right) + C$$
Simplify:
$$-\frac{1}{2}\left(\cos(u) – \frac{\cos^3(u)}{3}\right) + C= -\frac{1}{2}\cos(u) + \frac{1}{6}\cos^3(u) + C$$
Now substitute $u = 2x + 1$:
$$\boxed{\; \int \sin^3(2x + 1)\; dx = -\frac{1}{2}\cos(2x + 1) + \frac{1}{6}\cos^3(2x + 1) + C\;}$$
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NCERT Question 5: Evaluate the Integral
$$\displaystyle \int \sin^3(x)\cos^3(x)\; dx$$
Solution :
We know the power-reduction identity:
$$\sin^3(x)\cos^3(x) = (\sin(x)\cos(x))^3 = \left(\frac{1}{2}\sin(2x)\right)^3 = \frac{1}{8}\sin^3(2x)$$
Hence,
$$\int \sin^3(x)\cos^3(x)\; dx = \frac{1}{8} \int \sin^3(2x)\; dx$$
Let $u = 2x \Rightarrow du = 2\; dx \Rightarrow dx = \frac{du}{2}$
So,
$$\frac{1}{8} \int \sin^3(2x)\; dx = \frac{1}{16} \int \sin^3(u)\; du$$
Now use $\sin^3(u) = \sin(u)(1 – \cos^2(u))$:
$$\frac{1}{16} \int \sin^3(u)\; du = \frac{1}{16} \int \sin(u)(1 – \cos^2(u))\; du$$
Let $\cos(u) = t \Rightarrow -\sin(u)\; du = dt$
Then,
$$\frac{1}{16} \int \sin(u)(1 – \cos^2(u))\; du = -\frac{1}{16} \int (1 – t^2)\; dt$$
Integrate:
$$-\frac{1}{16} \left(t – \frac{t^3}{3}\right) + C$$
Substitute back $t = \cos(u)$:
$$-\frac{1}{16} \left(\cos(u) – \frac{\cos^3(u)}{3}\right) + C$$
Simplify:
$$= -\frac{1}{16}\cos(u) + \frac{1}{48}\cos^3(u) + C$$
Now substitute $u = 2x$:
$$\boxed{\displaystyle \int \sin^3(x)\cos^3(x)\; dx = -\frac{1}{16}\cos(2x) + \frac{1}{48}\cos^3(2x) + C}$$
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NCERT Question 6: Evaluate the Integral
$$\displaystyle \int \sin(x)\sin(2x)\sin(3x)\; dx$$
Solution
$$\displaystyle \int \sin(x)\sin(2x)\sin(3x)\; dx$$
We use the product-to-sum formula:
$$\sin A\sin B = \frac{1}{2}[\cos(A – B) – \cos(A + B)]$$
First, combine $\sin(x)\sin(3x)$:
$$\sin(x)\sin(3x) = \frac{1}{2}[\cos(2x) – \cos(4x)]$$
Now substitute into the integral:
$$\int \sin(x)\sin(2x)\sin(3x)\; dx = \frac{1}{2} \int [\cos(2x) – \cos(4x)]\sin(2x)\; dx$$
Expand:
$$= \frac{1}{2}\Bigl[\int \cos(2x)\sin(2x)\; dx – \int \cos(4x)\sin(2x)\; dx\Bigr]$$
(i) Evaluate $\displaystyle \int \cos(2x)\sin(2x)\; dx$
Use $\sin(2x)\cos(2x) = \dfrac{1}{2}\sin(4x)$
$$\int \cos(2x)\sin(2x)\; dx = \frac{1}{2}\int \sin(4x)\; dx = -\frac{1}{8}\cos(4x)$$
(ii) Evaluate $\displaystyle \int \cos(4x)\sin(2x)\; dx$
Use $\sin A\cos B = \dfrac{1}{2}[\sin(A + B) + \sin(A – B)]$
$$\sin(2x)\cos(4x) = \frac{1}{2}[\sin(6x) – \sin(2x)]$$
So,
$$\int \cos(4x)\sin(2x)\; dx = \frac{1}{2}\int [\sin(6x) – \sin(2x)]\; dx$$
$$= \frac{1}{2}\left(-\frac{1}{6}\cos(6x) + \frac{1}{2}\cos(2x)\right) = -\frac{1}{12}\cos(6x) + \frac{1}{4}\cos(2x)$$
Now substitute results back:
$$\int \sin(x)\sin(2x)\sin(3x)\; dx = \frac{1}{2}\Bigl[-\frac{1}{8}\cos(4x) – \left(-\frac{1}{12}\cos(6x) + \frac{1}{4}\cos(2x)\right)\Bigr]$$
Simplify:
$$= \frac{1}{2}\left(-\frac{1}{8}\cos(4x) + \frac{1}{12}\cos(6x) – \frac{1}{4}\cos(2x)\right)$$
$$= -\frac{1}{16}\cos(4x) + \frac{1}{24}\cos(6x) – \frac{1}{8}\cos(2x) + C$$
Final Answer
$$\boxed{\displaystyle \int \sin(x)\sin(2x)\sin(3x)\; dx = -\frac{1}{8}\cos(2x) – \frac{1}{16}\cos(4x) + \frac{1}{24}\cos(6x) + C}$$
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NCERT Question 7: Evaluate the Integral
$$\displaystyle \int \sin(4x)\sin(8x)\; dx$$
Solution :
$$\displaystyle \int \sin(4x)\sin(8x)\; dx$$
We use the product-to-sum identity:
$$\sin A\sin B = \frac{1}{2}[\cos(A – B) – \cos(A + B)]$$
Substitute $A = 4x$ and $B = 8x$:
$$\sin(4x)\sin(8x) = \frac{1}{2}[\cos(4x – 8x) – \cos(4x + 8x)]$$
$$= \frac{1}{2}[\cos(-4x) – \cos(12x)]$$
Since $\cos(-4x) = \cos(4x)$,
$$\sin(4x)\sin(8x) = \frac{1}{2}[\cos(4x) – \cos(12x)]$$
Now integrate:
$$\int \sin(4x)\sin(8x)\; dx = \frac{1}{2}\int [\cos(4x) – \cos(12x)]\; dx$$
Integrate each term:
$$= \frac{1}{2}\left(\frac{1}{4}\sin(4x) – \frac{1}{12}\sin(12x)\right) + C$$
Simplify:
$$= \frac{1}{8}\sin(4x) – \frac{1}{24}\sin(12x) + C$$
Final Answer
$$\boxed{\displaystyle \int \sin(4x)\sin(8x)\; dx = \frac{1}{8}\sin(4x) – \frac{1}{24}\sin(12x) + C}$$
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