integrate these

Q.2) multiply and divide the expression with sin5x.
the expression becomes :∫ sin5x(cos8x - cos7x)/(sin5x+2cos5x.sin5x) dx

=∫2 sin5x/2 cos5x/2 (2 sin15x/2 sinx/2) / sin5x + sin10x dx

=∫2 sin5x/2 cos5x/2 (2 sin15x/2 sinx/2) / 2 sin15x/2 sin5x/2 dx

=∫2 sin5x/2 sinx/2 dx =∫ cos2x - cos3x dx = sin2x/2 - sin3x/3 + c.

Q1) The given integral is
I = \int (1-2\sin^2\theta)^{3/2}\cos\theta \ \mathrm d\theta
Set \sin\theta = z\quad\Rightarrow \ \cos\theta \mathrm d\theta =\mathrm dz%I = \int (1-2\sin^2\theta)^{3/2}\cos\theta \ \mathrm d\theta
So the given integral
I = \int (1-2z^2)^{3/2}\ \mathrm dz
Next set
1-2z^2 = z^2t^2
so that z^2=\dfrac{1}{2+t^2}\qquad\Rightarrow\ z\ \mathrm dz = -\dfrac{t\ \mathrm dt}{(2+t^2)^2}
We then have
(1-2z^2)^{3/2}\ \mathrm dz=-z^3t^3\dfrac{t\ \mathrm dt}{z(2+t^2)^2}=-\dfrac{z^2t^4\ \mathrm dt}{(2+t^2)^2}=-\dfrac{t^4\ \mathrm dt}{(2+t^2)^3}
Hence,
I=\int\dfrac{-t^4}{(t^2+2)^3}\ \mathrm dt
Using partial fractions, we can write
\dfrac{-t^4}{(t^2+2)^3}=-\dfrac{1}{t^2+2}+\dfrac{4}{(t^2+2)^2}-\dfrac{4}{(t^2+2)^3}
Hence,
I=-\int\dfrac{1}{t^2+2}\ \mathrm dt +4\int\dfrac{1}{(t^2+2)^2}\ \mathrm dt-4\int\dfrac{1}{(t^2+2)^3}\ \mathrm dt
Consider the integral
I_n=\int\dfrac{1}{(t^2+a^2)^n}\ \mathrm dt
Integrate by parts
I_n=\dfrac{t}{(t^2+a^2)^n}+2n\int\dfrac{t^2}{(t^2+a^2)^{n+1}}\ \mathrm dt
\Rightarrow\ I_n=\dfrac{t}{(t^2+a^2)^n}+2n\int\dfrac{t^2+a^2-a^2}{(t^2+a^2)^{n+1}}\ \mathrm dt
\Rightarrow\ I_n=\dfrac{t}{(t^2+a^2)^n}+2n\int\dfrac{1}{(t^2+a^2)^n}\ \mathrm dt-2na^2\int\dfrac{1}{(t^2+a^2)^{n+1}}\ \mathrm dt
\Rightarrow\ I_n=\dfrac{t}{(t^2+a^2)^n}+2nI_n-2na^2I_{n+1}
Hence,
I_{n+1}=\dfrac{t}{2na^2(t^2+a^2)^n}+\dfrac{2n-1}{2na^2}I_n\quad (n>0)
As such we get for
J_n=\int\dfrac{1}{(t^2+2)^n}\ \mathrm dt
the reduction formula
J_n=\dfrac{t}{4(n-1)(t^2+2)^{n-1}}+\dfrac{2n-3}{4(n-1)}J_{n-1}\qquad (n>1)
Since,
J_1=\int\dfrac{1}{t^2+2}\ \mathrm dt=\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{t}{\sqrt{2}}
Therefore,
J_2=\dfrac{t}{4(t^2+2)}+\dfrac{1}{4\sqrt{2}}\tan^{-1}\dfrac{t}{\sqrt{2}}
And
J_3=\dfrac{t}{8(t^2+2)^2}+\dfrac{3}{8}\left(\dfrac{t}{4(t^2+2)}+\dfrac{1}{4\sqrt{2}}\tan^{-1}\dfrac{t}{\sqrt{2}}\right)
Now plugging back everything and simplifying (a bad task), we ultimately obtain
\int\dfrac{-t^4}{(2+t^2)^3}\ \mathrm dt = \dfrac{1}{16}\left(\dfrac{2t(6+5t^2)}{(2+t^2)^2}-3\sqrt{2}\tan^{-1}\dfrac{t}{\sqrt{2}}\right)+C
You need to go to the original variable by back-substitution. I think you will be ultimately ending with
\int(\cos2\theta)^{\frac{3}{2}}\cos\theta\ \mathrm d\theta =\dfrac{3}{8\sqrt{2}}\sin^{-1}(\sqrt{2}\sin\theta)+\dfrac{1}{8}\sqrt{\cos2\theta}\ (2\sin\theta+\sin3\theta)+C

Isnt that a bit like nuking a fly when merely swatting it would do?

After setting t = sin θ

\int (1-2t^2)^{\frac{3}{2}} dt = \frac{1}{\sqrt 2} \int (1-y^2)^{\frac{3}{2}} dy \ [y = t\sqrt 2 ]

= \frac{1}{\sqrt 2} \int \cos^4 \phi \ d \phi \ \ [ \sin \phi \ = y]

= \frac{1}{\sqrt 2} \int \left[ \frac{3}{8} + \frac{\cos 4 \phi}{8} + \frac{\cos 2 \phi}{8} \right] \ d \phi

= \frac{1}{\sqrt 2} \left[ \frac{3 \phi}{8} + \frac{\sin 4 \phi}{32} + \frac{\sin 2 \phi}{16} \right] + C \ \ [\phi = \sin^{-1} (\sqrt 2 \sin \theta) ]

Thanks a lot abhisek, manish sir, bhatt sir and kaymant sir