Question

Find the integral of the function.​

Answer 1

EXPLANATION.

⇒ ∫dx/sin(x - a)cos(x - b)

As we know that,

Multiply and divide numerator and denominator by = cos(a - b).

⇒ 1/cos(a - b) ∫cos(a - b).dx/sin(x - a)cos(x - b).

We can write as,

⇒ (a - b) = (x - b) - (x - a).

⇒ (a - b) = x - b - x + a.

⇒ (a - b) = - b + a.

⇒ (a - b) = a - b.

⇒ 1/cos(a - b) ∫ cos[(x - b) - (x - a)].dx/sin(x - a)cos(x - b).

As we know that,

⇒ cos(a - b) = cos(a).cos(b) + sin(a).sin(b).

⇒ 1/cos(a - b) ∫ cos(x - a)cos(x - b) + sin(x - a)sin(x - b).dx/sin(x - a)cos(x - b).

⇒ 1/cos(a - b) ∫cos(x - a)cos(x - b).dx/sin(x - a)cos(x - b) + ∫sin(x - a)sin(x - b).dx/sin(x - a)cos(x - b).

⇒ 1/cos(a - b) ∫cos(x - a).dx/sin(x - a) + ∫sin(x - b).dx/cos(x - b).

⇒ 1/cos(a - b) ∫[cot(x - a) + tan(x - b)].dx

⇒ 1/cos(a - b) [㏒| sin(x - a)| - ㏒ | cos(x - b) |] + c.

⇒ 1/cos(a - b) [ ㏒ | sin(x - a)/sin(x - b) | + c.

                                                                                                                         

MORE INFORMATION.

Integration of trigonometric functions.

(1) = ∫dx/a + bsin²x.

(2) = ∫dx/a + bcos²x.

(3) = ∫dx/acos²x + b sin(x).cos(x) + csin²x

(4) = dx/(a sin(x) + cos(x))².

Divide numerator and denominator by cos²x in all such type of  integrals and then put tan(x) = t.

Answer 2

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \blue{ \bf \: \int \: cotx \: = log(sinx) + c}}$

$(2). \: \boxed{ \blue{ \bf \: \int \:tanx \: = - log(cosx) + c }}$

$(3). \: \boxed{ \blue{ \bf \: cos(x - y) = \: cosx \: cosy \: + \: sinx \: siny}}$

$\large\underline\purple{\bold{Solution :- }}$

$\rm :\implies\: \int \: \dfrac{dx}{sin(x - a)cos(x - b)}$

On multiply and divide by cos(a - b), we get

$\rm :\implies\: \int \: \dfrac{1}{cos(a - b)} \times \dfrac{cos(a - b)}{sin(x - a)cos(x - b)} dx$

$\rm :\implies\:\dfrac{1}{cos(a - b)} \int \: \dfrac{cos(a - b)}{sin(x - a)cos(x - b)} dx$

$\rm :\implies\:\dfrac{1}{cos(a - b )} \int \: \dfrac{cos(a - b + x - x)}{sin(x - a)cos(x - b)} dx$

$\rm :\implies\:\dfrac{1}{cos(a - b)} \int \: \dfrac{cos((x - b) - (x - a))}{sin(x - a)cos(x - b)} dx$

$\rm :\implies\:\dfrac{1}{cos(a - b)} \int \: \dfrac{cos(x - b)cos(x - a) + sin(x - b)sin(x - a)}{sin(x - a)cos(x - b)} dx$

$\rm :\implies\:\dfrac{1}{cos(a - b)} \int \: (cot(x - a) + tan(x - b)) dx$

$\rm :\implies\:\dfrac{1}{cos(a - b)} \bigg(log(sin(x - a)) \: - \: log(cos(x - b)) \bigg) + c$

$\rm :\implies\:\dfrac{1}{cos(a - b)} \bigg(log(\dfrac{sin(x - a)}{cos(x - b)} \bigg) + c$