Question-43418

Question Number 43418 by Raj Singh last updated on 10/Sep/18

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18

\int \frac{c o s x d x}{s i n x c o s x + 1}

\int \frac{2 c o s x}{2 c o s x s i n x + 2} d x

\int \frac{c o s x + s i n x + c o s x - s i n x}{1 + {(s i n x + c o s x)}^{2}} d x

\int \frac{c o s x - s i n x}{1 + {(s i n x + c o s x)}^{2}} d x + \int \frac{c o s x + s i n x}{2 + 2 s i n x c o s x} d x

\int \frac{d (s i n x + c o s x)}{1 + (s i n x + c o s x)} - \int \frac{(c o s x + s i n x) d x}{- 3 + 1 - 2 s i n x c o s x}

\int \frac{d (s i n x + c o s x)}{1 + {(s i n x + c o s x)}^{2}} - \int \frac{d (s i n x - c o s x)}{- 3 + {(s i n x - c o s x)}^{2}}

f o r m u l a \int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} {t a n}^{-} (\frac{x}{a})

\int \frac{d x}{x^{2} - b^{2}}

= \frac{1}{2 b} \int \frac{(x + b) - (x - b)}{(x + b) (x - b)} d x

= \frac{1}{2 b} [\int \frac{d x}{x - b} - \int \frac{d x}{x + b}]

= \frac{1}{2 b} l n ∣ \frac{x - b}{x + b} ∣

s o a n s i s

{t a n}^{- 1} (s i n x + c o s x) - \frac{1}{2 \sqrt{3}} l n ∣ \frac{s i n x - c o s x - \sqrt{3}}{s i n x - c o s x + \sqrt{3}} ∣ + c