निमिन्लिखित समाकलनो को ज्ञात कीजिये:
(i) `int sin^(3)x cos^(2) x dx` (ii) `int (sin x)/(sin(x+a))dx` (iii) `int(1)/(1+tan x)dx`

Explanation

(i) `int sin^(3) x cos^(2) x dx`
`=int sin^(2) x cos^(2) x sin x dx`
`=int (1-cos^(2) x) cos^(2) x sin x dx`
मान लीजिए `cos x=t`
`:. -sin x dx =dt`
`=-int (1-t^(2))t^(2) dt`
`=-int (t^(2)-t^(4)) dt`
`=-[t^(3)/3-t^(5)/5]+c`
`=-1/3 cos^(3) x+1/5 cos^(5) x+c`.
(ii) `int (sin x)/(sin (x+a)) dx`
`x+a=t`
`:. dx=dt`
`=int (sin (t-a))/(sin t) dt`
`=int (sin t cos a-cos t sin a)/(sin t) dt`
`=int (sin t cos a)/(sin t) dt-int (cos t sin a)/(sin t) dt`
`=cos a int dt - sin a int cot t dt`
`=t cos a- sin a[log|sin t|+c_(1)]`
`=(x+a) cos a -sin a [log |sin (x+a)|+c_(1)]`
`=x cos a + a cos a - sin a log |sin (x+a)|-c_(1) sin a`
`= x cos a- sin a log |sin (x+a)|+c`, जहाँ `c=-c_(1) sin a +a cos a`
(iii) `int (1)/(1+tan x) dx`
`=int (1)/(1+(sin x)/(cos x)) dx`
`=int (cos x dx)/(cos x + sin x)`
`=1/2 int ((cos x+sin x+cos x-sin x))/((cos x+sin x)) dx`
`=1/2 int (cos x+sin x)/(cos x +sin x) dx +1/2 int (cos x-sin x)/(cos x + sin x) dx`
`=1/2 int dx +1/2 int (cos x- sin x)/(cos x+sin x) dx`
माना `cos x + sinx =t`
`:. (-sin x+cos x) dx=dt`
`=1/2 xx x+c_(1)/2 +1/2 int (dt)/t`
`=x/2+c_(1)/2+1/2 log |t|+c_(2)/2`
`=x/2+c_(1)/2+1/2 log |cos x+ sin x|+c_(2)/2`
`=x/2+1/2 log |cos x+sin x|+c` जहाँ `c=c_(1)/2+c_(2)/2`