`(i) (-3, 0) = {x : x ∈ R, -3 < x < 0}` \(e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right ) \\ = e^{u} \left ( \frac{sin ^{2} \;\frac{u}{2} + cos ^{2} \;\frac{u}{2} + 2 sin \;\frac{u}{2} \;cos\; \frac{u}{2}}{2\; cos ^{2} \;\frac{u}{2}} \right ) \\ = \frac{e^{u} \left ( sin \;\frac{u}{2} + cos \;\frac{u}{2} \right ) ^{2}}{2 cos ^{2} \;\frac{u}{2}} \\ = \frac{1}{2} e^{u} \left ( \frac{sin \;\frac{u}{2} + cos \;\frac{u}{2}}{cos \;\frac{u}{2}} \right ) ^{2} \\ = \frac{1}{2} e^{u} \left [ tan \;\frac{u}{2} + 1 \right ] ^{2} \\\) \(= \frac{1}{2} e^{u} \left [1 + tan \;\frac{u}{2} \right ] ^{2} \\ = \frac{1}{2} e^{u} \left [ 1 + tan ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ = \frac{1}{2} e^{u} \left [ sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = \left [ \frac{1}{2} sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \; \; \; ….. (1) \\ Suppose,\; tan\; \frac{u}{2} = f (u) \; so \; f’ (u) = \frac{1}{2} sec ^{2} \;\frac{u}{2}\) As we know, \(\int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C\) Considering equation (1), we get, \[\int \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = e^{u} tan\; \frac{u}{2} + C\] \begin{align} I&=\int \frac{1}{1 + cot\; \theta} d\theta \\ &= \int \large\frac{1}{1 + \frac{cos\; \theta}{sin\; \theta}} d\theta \\ &= \int \frac{sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ &= \frac{1}{2} \int \frac{2 sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ &= \frac{1}{2} \int \frac{(sin\; \theta + cos\; \theta) + (sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ &= \frac{1}{2} \int 1 d\theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ &= \frac{1}{2} \theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ \end{align} \begin{align} \int_0^{\Large\frac\pi2} (\sqrt{\sin x}+\sqrt{\cos x})^{-4}\ dx&=\int_0^{\Large\frac\pi2}\frac1{(\sqrt{\cos x})^4\ (1+\sqrt{\tan x})^4}\ dx\\ &=\int_0^{\Large\frac\pi2}\frac{\sec^2x}{(1+\sqrt{\tan x})^4}\ dx\\ &\stackrel{\color{red}{[1]}}=\int_0^\infty\frac{du}{(1+\sqrt{u})^4}\\ &\stackrel{\color{red}{[2]}}=2\int_0^\infty\frac{t}{(1+t)^4}\ dt\\ &\stackrel{\color{red}{[3]}}=2\cdot\text{B}(2,2)\\ &=\large\color{blue}{\frac13}. \end{align} \begin{align} \int cos3x.sin2x \, dx &= \int \cos(2x+x) \sin2x \,\frac{a}{b} dx \\ &= \int [\cos2x\cos x - \sin2x\sin x ]\sin2x \, dx \\ &= \int \cos2x \cos x \sin2x - \sin^2 2x \sin x \, dx \\ &= \int (\cos^2 x - \sin^2 x) \cos x (2 \sin x \cos x) - (2 \sin x \cos x)^2 \sin x \, dx \\ &= \int 2 \cos^4 x \sin x - 2 \cos^2 x \sin^3 x - 4 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x - 6 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x \, dx - \int 6 \sin^3 x \cos^2 x \, dx \\ \end{align} \begin{align} \int cos3x.sin2x \, dx &= \int \cos(2x+x) \sin2x \,\frac{a}{b} dx \\ &= \int [\cos2x\cos x - \sin2x\sin x ]\sin2x \, dx \\ &= \int \cos2x \cos x \sin2x - \sin^2 2x \sin x \, dx \\ &= \int (\cos^2 x - \sin^2 x) \cos x (2 \sin x \cos x) - (2 \sin x \cos x)^2 \sin x \, dx \\ &= \int 2 \cos^4 x \sin x - 2 \cos^2 x \sin^3 x - 4 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x - 6 \sin^3 x \cos^2 x \, dx \\ &= \int 2 \cos^4 x \sin x \, dx - \int 6 \sin^3 x \cos^2 x \, dx \\ \end{align}

\begin{align}I&=\int sin^3x.cos^3x \ dx \\ &=\int sin x.sin^2x.cos^3x\,dx\end{align}

`=>I=int\ sin x.sin^2x.cos^3x\dx `

`=>I=int\ sin x.(1-cos^2x).cos^3x\dx `

`I=int\ sin x.(cos^3x-cos^5x)\dx `

माना `cosx=t `

`:.\-sinx\dx=dt `

`I=-int (t^3-t^5)dt `

`=- t^4/4+t^6/6+c`

`= 1/6cost^6x-1/4cos^4x\+c`