`int(sin(x-alpha))/(sin(x+alpha)) dx`

1.	`int(sin(x-alpha))/(sin(x+alpha)) dx`
Answer» Correct Answer - `xcos 2alpha-sin 2alpha.log\|sin(x+alpha)\|+C` `I = int(sin(x+alpha-2alpha))/(sin(x+alpha)) dx`. ` = int(sin(x+alpha - 2alpha))/(sin(x+alpha)) dx` `= int(sin(x+alpha)cos2alpha-cos(x+alpha)sin2alpha)/(sin(x+alpha)) dx = cos 2alpha int dx - sin 2alpha . int(cos(x+alpha))/(sin(x+alpha))` `= x cos 2alpha - sin 2alpha. log\|sin(x+alpha)\| + C`.

Discussion

(i) `int(6x^(5) - 2/(x^(4)) - 7x + 3/x - 5 + 4e^(x) + 7^(x))dx` (ii) `int (8-x+2x^(3) -6/(x^(3)) +2x^(-5) + 5x^(-1)) dx` , (iii) `int (x/a+a/x+x^(a) +a^(x) + ax) dx`
Evaluate `int(secx)/((secx+tanx)) dx`.
Evaluate `int ((cosx - 2cos 2alpha)/(cosx - cos alpha)) dx`.
Evaluate : (i) `int((4-5 cosx)/(sin^(2)x))dx` , (ii) `int((1-cos2x)/(1+cos2x)) dx` (iii) `int (1)/(sin^(2)x cos^(2)x) dx` , (iv) `int(cos2x)/(cos^(2)x sin^(2) x) dx`
(i) `int(1)/((1+cos2x))dx` , (ii) `int (1)/((1-cos2x)) dx`
`int sqrt(1+sin2x)dx`
`int tan^(-1) ((sin2x)/(1+cos 2x))dx`
`intsqrt[1+sin2x]dx`A. `sinx + cosx + C`B. `-sinx + cos x + C`C. `sin x - cos x + C`D. `-sin x - cos x + C`
Evaluate : `intsqrt(1-sin2x)dx`.
Evaluate : (i) `int3x^(2)dx` , (ii) `int2^((x+3)) dx`