Question

int e^x(1+sin x/1+cos x)​

Answer 1

Step-by-step explanation:

Here it Is your Answer e^x tanx/2 +C

Answer 2

EXPLANATION.

$\sf \implies \int e^{x} \bigg( \dfrac{1 + sinx}{1 + cosx} \bigg)dx$

As we know that,

We can write equation as,

Formula of,

⇒ sin²∅ + cos²∅ = 1.

⇒ sin²x/2 + cos²x/2 = 1.

Apply this formula in equation, we get.

$\sf \implies \int e^{x} \bigg(\dfrac{sin^{2}\dfrac{x}{2} + cos^{2}\dfrac{x}{2} + 2sin\dfrac{x}{2} cos\dfrac{x}{2} }{2cos^{2}\dfrac{x}{2} } \bigg)dx$

$\sf \implies \int e^{x} \dfrac{\bigg(sin\dfrac{x}{2} + cos\dfrac{x}{2}\bigg)^{2} }{\bigg(2cos^{2}\dfrac{x}{2} \bigg) } dx$

$\sf \implies \dfrac{1}{2} \int e^{x} \bigg(\dfrac{sin\dfrac{x}{2} }{cos\dfrac{x}{2} } + \frac{cos\dfrac{x}{2} }{cos\dfrac{x}{2} } \bigg)^{2} dx$

$\sf \implies \dfrac{1}{2}\int e^{x} \bigg[tan\dfrac{x}{2} + 1 \bigg]^{2}dx$

$\sf \implies \dfrac{1}{2}\int e^{x} \bigg[ tan^{2} \dfrac{x}{2} + 1 + 2tan\dfrac{x}{2} \bigg]dx$

$\sf \implies \dfrac{1}{2} \int e^{x} \bigg[sec^{2}\dfrac{x}{2} + 2tan\dfrac{x}{2} \bigg]dx$

$\sf \implies \int e^{x} \bigg[\dfrac{1}{2} sec^{2} \dfrac{x}{2} + \dfrac{1}{2} \times 2tan\dfrac{x}{2}\bigg]dx$

$\sf \implies \int e^{x} \bigg[\dfrac{1}{2} sec^{2}\dfrac{x}{2} + tan\dfrac{x}{2} \bigg]dx$

As we know that,

Formula of : ∫eˣ [f(x) + f'(x)]dx.

⇒ ∫eˣ [f(x) + f'(x)]dx. = eˣf(x) + c.

Apply this formula in equation, we get.

$\sf \implies \int e^{x}\bigg[tan\dfrac{x}{2} + \dfrac{1}{2} sec^{2} \dfrac{x}{2} \bigg]dx.$

$\sf \implies e^{x} \ tan\dfrac{x}{2} + c$

                                                                                                                           

MORE INFORMATION.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k. dx = kx + c, (k∈R).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ 1).

(5) = ∫1/x dx = ㏒ₐx + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣ dx = aˣ/㏒(a) + c  Or  aˣ㏒ₐ(e) + c.