If I =(\integral \frac { d x } { \sec x + \operatorname { co... | MATH - INTEGRATION BY SUBSTITUTION | SolveeIt

If I = $\int \frac { d x } { \sec x + \operatorname { cosec } x ^ { \prime } }$ , then I equals:

$\frac { 1 } { 2 }$ $\left( \cos x + \sin x - \frac{1}{\sqrt{2}}\log(\cos ecx–\cos x) \right)$ + C

$\frac{1}{2}$ $\left( \sin x - \cos x - \frac { 1 } { \sqrt { 2 } } \log | \operatorname { cosec } x + \cot x | \right)$ + C

$\frac{1}{\sqrt{2}}$ $\left( \sin x + \cos x + \frac { 1 } { 2 } \log | \operatorname { cosec } x - \cos x | \right)$ + C

None of these

Answer

None of these

Explanation

I = $\int \frac { \sin x \cos x } { \sin x + \cos x }$ dx

= $\frac { 1 } { 2 }$ $\int_{}^{}\frac{(\sin x + \cos x)^{2} - 1}{\sin x + \cos x}$ dx

= $\frac { 1 } { 2 }$ $\int_{}^{}\left\lbrack \sin x + \cos x - \frac{1}{\sqrt{2}\sin(x + \pi/4)} \right\rbrack$ dx

= $\frac{1}{2}$ [sin x – cos x] – $\frac{1}{\sqrt{2}}$ log |cosec (x + p/4)

– cot (x + p/4)| + C

Question: If I =\(\int \frac { d x } { \sec x + \operatorname { cosec } x ^ { \prime } }\), then I equals:...