\displaystyle \textbf{Question 1: }~\int \sin 4x \,\cos 7x\,dx. \hspace{5.0cm} \;[\text{CBSE 2007}]
\displaystyle \text{Answer:}

\displaystyle \int \sin 4x \cos 7x\,dx

\displaystyle =\frac{1}{2}\int 2\cos 7x \sin 4x\,dx

\displaystyle =\frac{1}{2}\int \left[\sin(7x+4x)-\sin(7x-4x)\right]\,dx

\displaystyle =\frac{1}{2}\int \left(\sin 11x-\sin 3x\right)\,dx

\displaystyle =\frac{1}{2}\left[-\frac{\cos 11x}{11}+\frac{\cos 3x}{3}\right]+C

\displaystyle =-\frac{\cos 11x}{22}+\frac{\cos 3x}{6}+C

\displaystyle \textbf{Question 2: }~\int \cos 3x \,\cos 4x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \cos 4x \cos 3x\,dx

\displaystyle =\frac{1}{2}\int 2\cos 4x \cos 3x\,dx

\displaystyle =\frac{1}{2}\int \left[\cos(4x+3x)+\cos(4x-3x)\right]\,dx

\displaystyle =\frac{1}{2}\int (\cos 7x+\cos x)\,dx

\displaystyle =\frac{1}{2}\left[\frac{\sin 7x}{7}+\sin x\right]+C

\displaystyle =\frac{1}{14}\sin 7x+\frac{1}{2}\sin x+C

\displaystyle \textbf{Question 3: }~\int \cos(mx)\,\cos(nx)\,dx,\; m\ne n.
\displaystyle \text{Answer:}

\displaystyle \int \cos mx \cos nx\,dx

\displaystyle =\frac{1}{2}\int 2\cos(mx)\cos(nx)\,dx

\displaystyle =\frac{1}{2}\int \left[\cos\{(m+n)x\}+\cos\{(m-n)x\}\right]\,dx

\displaystyle =\frac{1}{2}\left[\frac{\sin (m+n)x}{m+n}+\frac{\sin (m-n)x}{m-n}\right]+C

\displaystyle \textbf{Question 4: }~\int \sin(mx)\,\cos(nx)\,dx,\; m\ne n.
\displaystyle \text{Answer:}

\displaystyle \int \sin(mx)\cos(nx)\,dx

\displaystyle =\frac{1}{2}\int 2\sin(mx)\cos(nx)\,dx

\displaystyle =\frac{1}{2}\int \left[\sin\{(m+n)x\}+\sin\{(m-n)x\}\right]\,dx

\displaystyle =\frac{1}{2}\left[-\frac{\cos (m+n)x}{m+n}-\frac{\cos (m-n)x}{m-n}\right]+C

\displaystyle \textbf{Question 5: }~\int \sin 2x \,\sin 4x \,\sin 6x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin 2x\,\sin 4x\,\sin 6x\,dx

\displaystyle =\frac{1}{2}\int (2\sin 2x\sin 4x)\sin 6x\,dx

\displaystyle =\frac{1}{2}\int [\cos(2x-4x)-\cos(2x+4x)]\sin 6x\,dx

\displaystyle =\frac{1}{2}\int [\cos 2x-\cos 6x]\sin 6x\,dx

\displaystyle =\frac{1}{2}\left[\int \cos 2x\sin 6x\,dx-\int \cos 6x\sin 6x\,dx\right]

\displaystyle =\frac{1}{4}\left[\int 2\cos 2x\sin 6x\,dx-\int 2\cos 6x\sin 6x\,dx\right]

\displaystyle =\frac{1}{4}\left[\int (\sin 8x+\sin 4x)\,dx-\int \sin 12x\,dx\right]

\displaystyle =\frac{1}{4}\left[-\frac{\cos 8x}{8}-\frac{\cos 4x}{4}+\frac{\cos 12x}{12}\right]+C

\displaystyle =-\frac{\cos 8x}{32}-\frac{\cos 4x}{16}+\frac{\cos 12x}{48}+C

\displaystyle \textbf{Question 6: }~\int \sin x \,\cos 2x \,\sin 3x\,dx.
\displaystyle \text{Answer:}

\displaystyle \int \sin x\,\cos 2x\,\sin 3x\,dx

\displaystyle =\frac{1}{2}\int (2\sin x\cos 2x)\sin 3x\,dx

\displaystyle =\frac{1}{2}\int \left[\sin(x+2x)+\sin(x-2x)\right]\sin 3x\,dx

\displaystyle =\frac{1}{2}\int (\sin 3x-\sin x)\sin 3x\,dx

\displaystyle =\frac{1}{2}\left[\int \sin^2 3x\,dx-\int \sin x\sin 3x\,dx\right]

\displaystyle =\frac{1}{4}\left[\int (1-\cos 6x)\,dx-\int \{\cos(x-3x)-\cos(x+3x)\}\,dx\right]

\displaystyle =\frac{1}{4}\left[\int dx-\int \cos 6x\,dx-\int \cos 2x\,dx+\int \cos 4x\,dx\right]

\displaystyle =\frac{1}{4}\left[x-\frac{\sin 6x}{6}-\frac{\sin 2x}{2}+\frac{\sin 4x}{4}\right]+C

\displaystyle =\frac{x}{4}-\frac{\sin 6x}{24}-\frac{\sin 2x}{8}+\frac{\sin 4x}{16}+C

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