How do i solve this integral?

[delicate veldt]

this looks interesting

ok try using the identity $\cos(2x)=\cos^{2}(x)-\sin^{2}(x)$

[vocal summit]

chudcel

[foggy axle]

Id use sin^2 (x)=(1-cos(2x))/2

[wet wolf]

so then its 1/4 integral os cos2x-cos^2(2x)

then use cos^2x = 1/2(1+cos2x)?

it should be a sin4x/8

?

[glass kindle]

for the first one with the cos2x, that +sin4x should be a -sin4x

your answer for the second integral looks way too complicated

i got a very simple answer

wait, what is all those fractions at the very bottom?

i used a different method for the second one, and got $\left(\frac{\cos4x}{32}-\frac{\cos2x}{8}\right)$

[vocal summit]

Helcovich Emire

[glass kindle]

but i checked it, and both are correct - they differ by a constant

though i think we need a +C at the end

[wet wolf]

Im solving for yp

We had to use the parameter method

[glass kindle]

what is the parameter method

[wet wolf]

https://prnt.sc/zAjgKDa44I6E

[wet wolf]

,solved

.solved