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प्रश्न
Evaluate the following integrals using properties of integration:
`int_0^(sin^2x) sin^-1 sqrt("t") "dt" + int_0^(cos^2x) cos^-1 sqrt("t") "dt"`
उत्तर
I1 = `int_0^(sin^2x) sin^-1 sqrt("t") "dt"`
Put `si^-1 sqrt("'t") = theta`
`sqrt("t")` = sin θ
t | 0 | sin2x |
θ | 0 | x |
`1/(2sqrt("t")) "dt"` = cos θ dθ
dt = `2sqrt("t") cos theta "d"theta`
= 2 sin θ cos θ dθ
dt = sin 2θ dθ
I1 = `int_0^x theta sin2theta "d"theta`
= `[ (- thetacos2theta)/2 + (sin2theta)/4]_0^x`
= `(-x cos 2x)/2 + (sin2x)/4` ......(1)
I1 = `int_0^(cos^2x) cos^-1 sqrt("t") "dt"`
Put `cos^-1 sqrt("t")` = θ
`sqrt("t"")` = cos θ
`1/(2sqrt("t")) "dt"` = – sin θ dθ
dt`- 2sqrt("t") sin theta "d"theta`
= – 2cos θ sin θ dθ
dt = – sin 2θ dθ
I2 = `int_0^(cos^x) cos^-1 sqrt("t") "dt"`
= `int_(pi/2)^x - theta sin 2theta "d"theta`
t | 0 | cos2x |
θ | `pi/2` | x |
= `- [(- theta cos 2theta)/2+ (sin theta)/4]_(pi/2)^x`
= `[(theta cos2theta)/2 - (sin 2theta)/4]_(p/2)^x`
= `(x cos 2x)/2 - (sin 2x)/4 + pi/4` ........(2)
I = I1 + I2
= `(- x cos 2x)/2 + (sin 2x)/4 + (x cos2x)/2 - (sin 2x)/4 + pi/4`
I = `pi/4`
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