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Question Number 59631 by Mr X pcx last updated on 12/May/19

1) c a l c u l a t e \int_{0}^{2 π} \frac{d x}{a c o s x + b s i n x}

w i t h a, b r e a l s

2) f i n d a l s o \int_{0}^{2 π} \frac{c o s x d x}{{(a c o s x + b s i n x)}^{2}} a n d

\int_{0}^{2 π} \frac{s i n x d x}{{(a c o s x + b s i n x)}^{2}}

3) f i n d t h e v a l u e o f \int_{0}^{2 π} \frac{d x}{2 c o s x + \sqrt{3} s i n x}

Commented by Mr X pcx last updated on 12/May/19

4) f i n d t h e v a l u e o f \int_{0}^{2 π} \frac{c o s x d x}{{(3 c o s x + \sqrt{3} s i n x)}^{2}}

Commented by maxmathsup by imad last updated on 15/May/19

w e h a v e I = \int_{0}^{π} \frac{d x}{a c o s x + b s i n x} + \int_{π}^{2 π} \frac{d x}{a c o s x + b s i n x} = H + K

H =_{t a n (\frac{x}{2}) = t} \int_{0}^{\infty} \frac{1}{a \frac{1 - t^{2}}{1 + t^{2}} + b \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{0}^{\infty} \frac{2 d t}{a - {a t}^{2} + 2 b t}

= \int_{0}^{\infty} \frac{- 2 d t}{{a t}^{2} - 2 b t - a} r o o t s o f {a t}^{2} - 2 b t - a = 0

Δ^{^{'}} = b^{2} + a^{2} > 0 f o r a \neq 0 \Rightarrow t_{1} = \frac{b + \sqrt{a^{2} + b^{2}}}{a}

t_{2} = \frac{b - \sqrt{a^{2} + b^{2}}}{a} \Rightarrow H = \int_{0}^{\infty} \frac{- 2}{a (t - t_{1}) (t - t_{2})} d t = \frac{- 2}{a (t_{1} - t_{2})} \int_{0}^{\infty} (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) d t

= \frac{- 2}{\frac{2 a \sqrt{a^{2} + b^{2}}}{a}} {[l n ∣ \frac{t - t_{1}}{t - t_{2}} ∣]}_{0}^{+ \infty} = \frac{- 1}{\sqrt{a^{2} + b^{2}}} (- l n ∣ \frac{t_{1}}{t_{2}} ∣) = \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{b + \sqrt{a^{2} + b^{2}}}{b - \sqrt{a^{2} + b^{2}}} ∣

\Rightarrow H = \frac{1}{\sqrt{a^{2} + b^{2}}} l n (\frac{\sqrt{a^{2} + b^{2}} + b}{\sqrt{a^{2} + b^{2}} - b})

K =_{x = t + π} \int_{0}^{π} \frac{d t}{- a c o s t - b s i n t} = - \frac{1}{\sqrt{a^{2} + b^{2}}} l n (\frac{\sqrt{a^{2} + b^{2}} + b}{\sqrt{a^{2} + b^{2}} - b}) \Rightarrow

I = 0

Commented by maxmathsup by imad last updated on 15/May/19

2) l e t f (a) = \int_{0}^{2 π} \frac{d x}{a c o s x + b s i n x} \Rightarrow f^{^{'}} (a) = - \int_{0}^{2 π} \frac{c o s x}{{(a c o s x + b s i n x)}^{2}} d x

b u t f (a) = 0 \Rightarrow f^{^{'}} (a) = 0 \Rightarrow \int_{0}^{2 π} \frac{c o s x}{{(a c o s x + b s i n x)}^{2}} d x = 0 a l s o w e h a v e

\int_{0}^{2 π} \frac{s i n x}{{(a c o s x + b s i n x)}^{2}} d x = 0

3) \int_{0}^{2 π} \frac{d x}{2 c o s x + \sqrt{3} s i n x} = 0 .

Answered by tanmay last updated on 12/May/19

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

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

\frac{1}{r} \int \frac{d x}{s i n (α + x)}

\frac{1}{r} \int c o s e c (x + α) d x

\frac{1}{r} \times l n t a n (\frac{x + α}{2}) + c

\frac{1}{\sqrt{a^{2} + b^{2}}} l n t a n (\frac{x + {t a n}^{- 1} (\frac{a}{b})}{2}) + c

n o w p r o c e e d

Answered by tanmay last updated on 13/May/19

2) \int \frac{c o s x}{{(a c o s x + b s i n x)}^{2}}

c o s x = A (a c o s x + b s i n x) + B \times \frac{d}{d x} (a c o s x + b s i n x)

c o s x = A (a c o s x + b s i n x) + B \times (- a s i n x + b c o s x)

c o s x = (A a + b B) c o s x + (A b - B a) s i n x

A b - B a = 0

\frac{A}{a} = \frac{B}{b} = k \to A = a k a n d B = b k

A a + B b = 1

a^{2} k + b^{2} k = 1

k = \frac{1}{a^{2} + b^{2}} s o A = \frac{a}{a^{2} + b^{2}} a n d B = \frac{b}{a^{2} + b^{2}}

n o w

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

\int [\frac{A (a c o s x + b s i n x) + B \times \frac{d}{d x} \frac{(a c o s x + b s i n x)}{}}{{(a c o s x + b s i n x)}^{2}}] d x

\int \frac{A}{(a c o s x + b s i n x)} d x + B \int \frac{d (a c o s x + b s i n x)}{{(a c o s x + b s i n x)}^{2}}

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

n o w p l s p u t t h e u p p e r a n d l o w e r l i m i t

f o r \int \frac{s i n x d x}{{(a c o s x + b s i n x)}^{2}} \to s a m e m e t h o d \dots

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