Advertisements
Advertisements
प्रश्न
Evaluate:
`int_(-pi/4)^(pi/4) (1)/(1 - sinx)*dx`
मूल्यांकन
Advertisements
उत्तर
`int_(-pi/4)^(pi/4) (1)/(1 - sinx)*dx`
= `int_(-pi/4)^(pi/4) (1)/(1 - sinx)*(1 + sinx)/(1 + sinx)*dx`
= `int_(-pi/4)^(pi/4)(1 + sinx)/(1 - sin^2x)*dx`
= `int_(-pi/4)^(pi/4)(1 + sinx)/(cos^2x)*dx`
= `int_(-pi/4)^(pi/4) 1/(cos^2x) dx + int_(-pi/4)^(pi/4) (sinx)/(cos^2x)*dx`
= `int_(-pi/4)^(pi/4)sec^2x dx + int_(-pi/4)^(pi/4) 1/cosx*sinx/cosxdx`
= `[tanx]_(-pi/4)^(pi/4) + int_(-pi/4)^(pi/4) secx tanx*dx`
= `[tan pi/4 - tan (-pi/4)] + [secx]_(-pi/4)^(pi/4)`
= `[1 + 1] + [sec pi/4 - sec ((-pi)/4)]`
= `2 + [sqrt2 - sqrt2]`
= 2 + 0
I = 2
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
