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Question Number 64017 by Rio Michael last updated on 12/Jul/19

w h y {c a n}^{'} t w e d i f f e r e n t i a t e o r i n t e r g r a t e p o w e r s o f t r i g o n o m e t r i c

f u n c t i o n s s u c h a s

1) \int {c o s}^{2} x d x ? 3) \int {t a n}^{2} 2 x d x

2) \int {s i n}^{2} x d x ? 4) \int {s i n}^{10} x

h e n c e h o w d o w e s o l v e s u c h p r o b l e m s . ?

Commented by kaivan.ahmadi last updated on 12/Jul/19

c o s 2 x = 2 {c o s}^{2} x - 1 \Rightarrow {c o s}^{2} x = \frac{1 + c o s 2 x}{2}

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

\frac{1}{2} (x + \frac{1}{2} s i n 2 x) + C

Commented by kaivan.ahmadi last updated on 12/Jul/19

c o s 2 x = 1 - 2 {s i n}^{2} x \Rightarrow {s i n}^{2} x = \frac{1 - c o s 2 x}{2}

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

\frac{1}{2} (x - \frac{1}{2} s i n 2 x) + C

Commented by kaivan.ahmadi last updated on 12/Jul/19

\int {t a n}^{2} 2 x d x = \int (1 + {t a n}^{2} 2 x - 1) d x = \frac{1}{2} t a n 2 x - x + C

Commented by MJS last updated on 12/Jul/19

look at qu . 64037

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