integrate sinx+cosx / 9+16sin2x with respect to x
Answers
EXPLANATION.
⇒ ∫sin(x) + cos(x)/ 9 + 16sin2x dx.
Put = (sin(x) - cos(x)) = t.
Differentiate both sides w.r.t t, we get.
⇒ [cos(x) + sin(x)]dx = dt.
Square both the sides, we get.
⇒ [cos(x) - sin(x)]² = t².
⇒ [cos²x + sin²x - 2sinxcosx] = t².
⇒ [1 - 2sinxcosx] = t².
As we know that,
Sin(2x) = 2sinxcosx.
⇒ [1 - sin2x] = t².
⇒ sin2x = 1 - t².
Put the values in equation, we get.
⇒ ∫sin(x) + cos(x)dx/9 + 16sin2x.
⇒ ∫dt/9 + 16(1 - t²).
⇒ ∫dt/9 + 16 - 16t².
⇒ ∫dt/25 - 16t².
⇒ ∫dt/16(25/16 - t²).
⇒ 1/16∫dt/25/16 - t².
⇒ 1/16∫dt/(5/4)² - t².
As we know that,
⇒ ∫dx/a² - x² = 1/2a㏒(a + x/a - x) + c.
⇒ 1/16 [ 1/2(5/4)㏒(5/4 + t / 5/4 - t)] + c.
⇒ 1/16 [ 2/5㏒(5 + 4t/4/5 - 4t/4) ]+ c.
⇒ 1/16 [ 2/5㏒(5 + 4t/5 - 4t)] + c.
Put the value of t in equation,
⇒ 1/16 [ 2/5㏒(5 + 4(sin x + cos x)/5 - 4(sin x - cos x)] + c.
MORE INFORMATION.
⇒ ∫0.dx = c.
⇒ ∫1.dx = x + c.
⇒ ∫kdx = kx + c, (K∈R).
⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ - 1).
⇒ ∫1/x (dx) = ㏒x + c.
⇒ ∫eˣdx = eˣ + c.
⇒ ∫aˣdx = aˣ/㏒a + c = aˣ㏒e + c.