Question

Solve this integral pleaseee​

Answer 1

Answer:

Step-by-step explanation:

lets take sin theta =t

cos theta dtheta = dt

∫3t-2 dt / 4+t^2-4t

3t-2 can be written as

lamda(2t-4)+meu

lamda = 3/2

-4lamda+meu=-2

-6+meu=-2

meu=4

therefore 3t-2 = 3/2(2t-4)+4

∫3/2(2t-4)+4 dt / t^2-4t+4

3/2∫2t-4 dt /t^2-4t+4 + 4∫dt/t^2-4t+4

3/2 ln t^2-4t+4) + 4∫dt/(t-2)^2

3/2 ln t^2-4t+4) + 4(t-2)^-1/-1

3/2 ln t^2-4t+4 ) -4/t-2

3/2 ln (sin^2 x-4sinx+4) -4/sinx-2