Question

Please solve this ASAP
and please no fake answers
​

Answer 1

EXPLANATION.

$\sf \implies \int \dfrac{sin2xdx}{acos^{2}x + bsin^{2} x }$

Using the formula ⇒ cos²x = 1 - sin²x.

$\sf \implies \int \dfrac{Sin2xdx}{a(1 - sin^{2}x )+ bsin^{2} x }$

$\sf \implies \int \dfrac{Sin2xdx}{a - asin^{2}x + bsin^{2}x }$

$\sf \implies \int\dfrac{Sin2xdx}{a + sin^{2}x(b - a) }$

Using the formula ⇒ Sin2x = 2Sinx.Cosx.

$\sf \implies \int \dfrac{(2Sinx.Cosx)dx}{a + sin^{2}x(b - a) }$

By using the Substitution method, we get.

Substitute = a + (sin²x(b - a)) = t.

Differentiate w.r.t x, we get.

⇒ 0 + (b - a)(2SinxCosx)dx = dt.

⇒ (2Sinx.Cosx)dx = dt/(b - a).

Substitute the value in equation, we get.

$\sf \implies \int \dfrac{dt}{t(b - a)}$

As we know that,

⇒ (b - a) is a constant term,

$\sf \implies \dfrac{1}{(b - a)} \int \dfrac{dt}{t}$

As we know that,

⇒ ∫dt/t = ㏒ | t | + c.

$\sf \implies \dfrac{1}{(b - a)} log | t | + c$

Put the value of t in equation, we get.

$\sf \implies \dfrac{1}{(b - a)} log | a + (b - a)sin^{2}x | + c$

                                                                                           

MORE INFORMATION.

Integration of Trigonometric functions,

(1) = ∫P(sin(x)) + q.(cos(x))dx/a sin(x) + b cos(x).

(2) = ∫p(sin(x))dx/a sin(x) + b cos(x).

(3) = ∫q cos(x)/a sin(x) + b cos(x).

For their integration, we first express Nr. as follows,

Nr = A(Dr) + B(derivative of Dr).

Then integral = Ax + B ㏒(Dr) + c.

Answer 2

Answer:

EXPLANATION.

\sf \implies \int \dfrac{sin2xdx}{acos^{2}x + bsin^{2} x }⟹∫

acos

2

x+bsin

2

x

sin2xdx

Using the formula ⇒ cos²x = 1 - sin²x.

\sf \implies \int \dfrac{Sin2xdx}{a(1 - sin^{2}x )+ bsin^{2} x }⟹∫

a(1−sin

2

x)+bsin

2

x

Sin2xdx

\sf \implies \int \dfrac{Sin2xdx}{a - asin^{2}x + bsin^{2}x }⟹∫

a−asin

2

x+bsin

2

x

Sin2xdx

\sf \implies \int\dfrac{Sin2xdx}{a + sin^{2}x(b - a) }⟹∫

a+sin

2

x(b−a)

Sin2xdx

Using the formula ⇒ Sin2x = 2Sinx.Cosx.

\sf \implies \int \dfrac{(2Sinx.Cosx)dx}{a + sin^{2}x(b - a) }⟹∫

a+sin

2

x(b−a)

(2Sinx.Cosx)dx

By using the Substitution method, we get.

Substitute = a + (sin²x(b - a)) = t.

Differentiate w.r.t x, we get.

⇒ 0 + (b - a)(2SinxCosx)dx = dt.

⇒ (2Sinx.Cosx)dx = dt/(b - a).

Substitute the value in equation, we get.

\sf \implies \int \dfrac{dt}{t(b - a)}⟹∫

t(b−a)

dt

As we know that,

⇒ (b - a) is a constant term,

\sf \implies \dfrac{1}{(b - a)} \int \dfrac{dt}{t}⟹

(b−a)

1

∫

t

dt

As we know that,

⇒ ∫dt/t = ㏒ | t | + c.

\sf \implies \dfrac{1}{(b - a)} log | t | + c⟹

(b−a)

1

log∣t∣+c

Put the value of t in equation, we get.

\sf \implies \dfrac{1}{(b - a)} log | a + (b - a)sin^{2}x | + c⟹

(b−a)

1

log∣a+(b−a)sin

2

x∣+c