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Question Number 93098 by Rio Michael last updated on 10/May/20

find the general solution to

\int \frac{1}{a \sin x + b \cos x} d x and \int \frac{1}{a \cos x - b \sin x} d x

where a, b are constants .

Commented by prakash jain last updated on 10/May/20

a \sin x + b \cos x

= \sqrt{a^{2} + b^{2}} (\frac{a}{\sqrt{a^{2} + b^{2}}} \sin x + \frac{b}{\sqrt{a^{2} + b^{2}}} \cos x)

\frac{a}{\sqrt{a^{2} + b^{2}}} = \cos α

\frac{b}{\sqrt{a^{2} + b^{2}}} = \sin α

\sqrt{a^{2} + b^{2}} = \sqrt{a^{2} + b^{2}} \sin (α + x)

\frac{1}{a \sin x + b \cos x} = \sqrt{a^{2} + b^{2}} cosec (x + α)

now use cosec formula for integration .

- - - - - -

Same procedure can be use

\frac{1}{a \sin x - b \cos x}

Commented by Rio Michael last updated on 10/May/20

thank you sir .

Commented by abdomathmax last updated on 11/May/20

I = \int \frac{d x}{a c o s x + b s i n x} w e s u p p o s e a \neq 0 \Rightarrow

I = \frac{1}{a} \int \frac{d x}{c o s x + λ s i n x} = A_{λ} (λ = \frac{b}{a})

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

A_{λ} = \frac{1}{a} \int \frac{2 d t}{(1 + t^{2}) (\frac{1 - t^{2}}{1 + t^{2}} + λ \frac{2 t}{1 + t^{2}})}

{a A}_{λ} = \int \frac{2 d t}{1 - t^{2} + 2 λ t} = - \int \frac{2 d t}{t^{2} - 2 λ t - 1}

l e t s o l v e t^{2} - 2 λ t - 1 = 0

Δ^{^{'}} = λ^{2} + 1 \Rightarrow t_{1} = λ + \sqrt{1 + λ^{2}}

t_{2} = λ - \sqrt{1 + λ^{2}}

{a A}_{λ} = - \int \frac{2 d t}{(t - t_{1}) (t - t_{2})}

= - \frac{2}{2 \sqrt{1 + λ^{2}}} \int (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) d t

= \frac{1}{\sqrt{1 + λ^{2}}} l n ∣ \frac{t - t_{2}}{t - t_{1}} ∣ + C

= \frac{1}{\sqrt{1 + λ^{2}}} l n ∣ \frac{t - λ + \sqrt{1 + λ^{2}}}{t - λ - \sqrt{1 + λ^{2}}} ∣ + C

= \frac{1}{\sqrt{1 + \frac{b^{2}}{a^{2}}}} l n ∣ \frac{t - \frac{b}{a} + \sqrt{1 + \frac{b^{2}}{a^{2}}}}{t - \frac{b}{a} - \sqrt{1 + \frac{b^{2}}{a^{2}}}} ∣ + C

= \frac{∣ a ∣}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{\frac{a t - b}{a} + \frac{\sqrt{a^{2} + b^{2}}}{∣ a ∣}}{\frac{a t - b}{a} - \frac{\sqrt{a^{2} + b^{2}}}{∣ a ∣}} ∣ + C

i f w e t a k e a > 0 w e g e t

I = \frac{1}{\sqrt{a^{2} + b^{2}}} l n ∣ \frac{a t a n (\frac{x}{2}) - b + \sqrt{a^{2} + b^{2}}}{a t a n (\frac{x}{2}) - b - \sqrt{a^{2} + b^{2}}} ∣ + C

=

=

Commented by Rio Michael last updated on 11/May/20

wow sir spendid

Commented by mathmax by abdo last updated on 11/May/20

t h a n k x