If ∫ d θ cos 2 θ tan 2 θ + sec 2 θ = λ tan θ + 2 log e f θ + C where C is a constant of integration, then the ordered pair λ , f θ is equal to:

Question

If &#8747; d &#952; cos 2 &#952; tan 2 &#952; + sec 2 &#952; = &#955; tan &#952; + 2 log e f &#952; + C where C is a constant of integration, then the ordered pair &#955; , f &#952; is equal to:

Aicharya · Accepted Answer

&#8747; sec 2 &#952; 1 + tan 2 &#952; 1 - tan 2 &#952; + 2 tan &#952; 1 - tan 2 &#952; d &#952; = &#8747; sec 2 &#952; 1 - tan 2 &#952; 1 + t a n &#952; 2 d &#952; = &#8747; sec 2 &#952; 1 - tan &#952; 1 + tan &#952; d &#952; Let tan &#952; = t &#8658; sec 2 &#952; d &#952; = d t = &#8747; 1 - t 1 + t d t = &#8747; - 1 + 2 1 + t d t = - t + 2 ln 1 + t + C = - tan &#952; + 2 ln 1 + tan &#952; + C &#8658; &#955; = - 1 &#160; a n d &#160; f &#952; = 1 + tan &#952;

If ∫ d θ cos 2 θ tan 2 θ + sec 2 θ = λ tan θ + 2 log e f θ + C…

Solution

If &#8747; d &#952; cos 2 &#952; tan 2 &#952; + sec 2 &#952; = &#955; tan &#952; + 2 log e f &#952; + C…

Solution

If ∫ d θ cos 2 θ tan 2 θ + sec 2 θ = λ tan θ + 2 log e f θ + C…