integration problems..........

Ans ) I am getting

\ln \left|1 + x \right|\ - 2 \left[\frac{1}{x^{2}}- \frac{\tan^{-1} (x^{2}-1)}{x^{2}} \right]

Is it correct ?

nihal ,the best way z the graphical approach,the graphical meaning of integration.we'll discuss it in class. Waise maine tmhe hint de diya hai

# 8

1] ∫dx1+tanx = ∫cosxsinx+cosxdx = ∫cosx+sinx + cosx - sinx 2(sinx+cosx)

= x2 + ∫ cosx - sinx dx2(sinx+cosx)

put sinx+cosx = t

2] ∫1+sinx-1 dx√1+sinx =∫√1+sinxdx - ∫1√1+sinx dx

now use 1+sinx = (sinx2+cosx2)²= 2sin²(x2+Ï€4)

3] put cosx = t

4]∫dxsinxcos³x = ∫sec⁴xtanx dx = ∫ (1+tan²x)sec²xtanxdx

put tanx = t

5]∫sec^3/4x cosec^5/4x dx = ∫dxsin^5/4x cos^3/4x =∫ dxsin^5/4xcos^5/4xcos^3/4xcos^5/4 = ∫ sec²xdxtan^5/4x

put tanx = t

#7

1] x³-1 = t³

x²dx = t²dt

so I = ∫ (x³-1)^1/3 x⁵ .dx = ∫t (t³+1)t²dt = ∫t⁶+t³ dt

2] e^x - 1 = t
e^xdx = dt
dx = dtt+1

I = ∫dtt(t+1)

3] e^x+1 = t

e^xdx = dt

dx = dtt-1

I = ∫t-2t(t+1) dt

4] put sin(x³) = t

1) ∫11+ tanx dx

2) ∫ sinx√1 + sinxdx

3) ∫ cos⁹xsinxdx

4) ∫ dxsinx.cos³x

5) ∫ sec^3/4x . cosec^5/4x.dx

Put tan √x=z thus dz=sec^2 (√x)[1/2√x]dx
Thus I=2∫z^4 dz
=2/5 z^5
=2/5 (tan^5(√x) ).(ans)