Integreation (1)

given is xⁿ

differentiation of xⁿ=nx^n-1

so think of bringing something of type x^n-1 in the numerator

for third try putting √x=t

for first

put cotx=t
-cosec²xdx=dt
-(1+t²)dx=dt
dx=-dt/(1+t²)
The integral becomes

(1-t^n-2)dt(1/t+t^n-1)(1+t²)

(t-t^n-1)dt(1+tⁿ)(1+t²)

try to separate the denominator you will get the answer

after getting
dt(1+t)√(1-t²)

just put t=sinθ and proceed

first one reduces to

tⁿ-t²t(1+tⁿ)(1+t²)=1t(1+t²)-1t(1+tⁿ)

now if you solved the second one then this can be solved in the same way