Integration

WORKING for the first question
1) I = ∫ [ 1- cot ^n-2 x ) / (tanx + cot ^n-1 x) ] dx
Now suppose, we put n=2, we get
I = 0 which satisfies R.H.S
But this method is correct or not???

WORKING for the second one,
2) I ₁ = ∫ sin ^-1 x dx
Put sint = x,
therefore, dx = cost dt
I ₁ = ∫ t cost dt = cos t ∫ t - ∫ (-sint) ∫t dt (By using by part)
I ₁ = cost (t²/2) + ∫ sint (t ²/2) = x sin ^-1 x + √1-x ² + c

For I 2 , put 1-x ² = t ²
√1-x ² = x
That implies, - 2x dx = 2t dt
Therefore, I ₂ = ∫ sin ^-1 t dt / √1-t ²
= ∫ [ sin ^-1t / √1-t ² ] dt
since, d(sin ^-1 t) / d t = 1 / √ 1-t²
Therefore, I ₂ = ∫ (sin ^-1 t) ² / 2 dt
But still I am not getting I ₁ + I ₂ = pie x /2
Plz help

WORKING for the third one,
I = ∫ [ cos ² 2x / cos 2x] dx

since 1+cos 2 (2x) = 2 cos ² (2x)
Therefore, on substituting,
we get, I = 1/2 ∫ [ 1 + cos 2(2x)] / cos 2x] dx
But restof it, I am not getting. Plz help..

Q.2) I₁=∫sin^-1x dx now,put integration by parts :

I₁ = sin^-1x . x - ∫x / √(1-x²) (substitute 1-x²=t and solve )
u'll get I₁ = sin^-1x.x - (1-x²)^1/2

now,same for I₂, put integration by parts.better write I₂ as cos^-1x.
u'll get I₂ = cos^-1x . x + (1-x²)^1/2

I₁+I₂ = pi x /2.

the method u approached is also right,but it would become lengthy.

need not integerate f I1 =∫ sin -1 x dx and I 2 = ∫ sin -1 √1-x 2 dx seperatly..
I₂=cos^-1x...so I₁+I₂=pi/2
∫pi/2 dx=pi x/2.