# please give me the solution for this question as soon as possible

Sumit Majumdar IIT Delhi
9 years ago
Solution:
I = ∫ dx / (sinx + secx)

= ∫ [ cosx / (cosxsinx + 1) ] dx

= ∫ [2cosx / 2(cosxsinx + 1) ] dx

= ∫ [ ( cosx + cosx + sinx - sinx) / (2cosxsinx + 2) ] dx

= ∫ [ ( cosx + sinx) / (2cosxsinx + 2) ] dx + ∫ [ ( cosx - sinx) / (2cosxsinx + 2) ] dx

= - ∫ [ (cosx + sinx)/( -3 - 2cosxsinx + cos²x + sin²x ) ] dx + ∫ [ ( cosx - sinx) / (2cosxsinx + 1 + cos²x + sin²x ) ] dx

= - ∫ [ (cosx + sinx)/( -3 + (cosx - sinx)² ) ] dx + ∫ [ ( cosx - sinx)/ ( 1 + ( cosx + sinx)² ) ] dx

= - ∫ [ 1 /( -3 + (sinx - cosx )² ) ] d(sinx - cosx) + ∫ [ 1/ ( 1 + (sinx + cosx)² ) ] d(sinx + cosx)

= - ∫ [ 1/( -3 + (sinx - cosx )² ) ] d(sinx - cosx) + ∫ [ 1/ ( 1 + (sinx + cosx)² ) ] d(sinx + cosx)

= - ∫ [ 1/( -3 + t² ) ] dt + ∫ [ 1/ ( 1 + u² ) ] du

= ( 1/√3 ) ln | (t + √3) / (t - √3) | + arctgu + C

= ( 1/√3 ) ln | (sinx - cosx + √3) / (sinx - cosx - √3) | + arctg(sinx + cosx) + C
Thanks & Regards
Sumit Majumdar,
Ph.D,IIT Delhi
Sumit Majumdar IIT Delhi
9 years ago
Dear student,
We can have the following result:
$\int \frac{dx}{sinx+secx}=\int \frac{cosx}{sin^{2}x+1}$
Now if I asume, u = sinx, so we get:
$\int \frac{cosx}{sin^{2}x+1}=\int \frac{du}{1+u^{2}}=tan^{-1}u=tan^{-1}\left (sinx \right )+c$
Regards
Sumit