integrate (sin^3x + cos^3x) dx
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Answer:
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EXPLANATION.
⇒ ∫(sin³x + cos³x) dx.
As we know that,
We can write equation as,
⇒ I = ∫(sin³x)dx + ∫(cos³x)dx.
⇒ I = I₁ + I₂.
⇒ I₁ = ∫(sin³x)dx.
⇒ I₁ = ∫(sin²x)(sin x)dx.
As we know that,
⇒ sin²x + cos²x = 1.
⇒ sin²x = 1 - cos²x.
Put the value of sin²x in equation, we get.
⇒ ∫(sin x)[1 - cos²x]dx.
⇒ ∫(sin x) - (sin x cos²x)dx.
⇒ ∫(sin x)dx - ∫(sin x cos²x)dx.
⇒ ∫(sin x)dx. = -cos x + c.
⇒ ∫sin x cos²x dx.
By using the substitution method, we get.
Let we assume that,
⇒ cos x = t.
Differentiate w.r.t x, we get.
⇒ -sin x dx = dt.
Put the value in equation, we get.
⇒ -∫t²dt.
As we know that,
⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c.
Using this formula in equation, we get.
⇒ -[t³/3] + c.
Put the value of t = cos x in equation, we get.
⇒ -[cos³x/3] + c.
⇒ I₁ = ∫(sin³x)dx = -cos x + cos³x/3 + c.
⇒ I₁ = ∫(sin³x)dx = cos³x/3 - cos(x) + c.
⇒ I₂ = ∫cos³xdx.
As we know that,
Formula of :
⇒ cos3x = 4cos³x - 3cosx.
⇒ cos3x + 3cosx = 4cos³x.
⇒ cos3x + 3cosx/4 = cos³x.
⇒ ∫cos3x + 3cosx/4 dx.
⇒ 1/4∫cos3x + 3cosx dx.
⇒ 1/4∫cos3x dx + ∫3cosx dx.
⇒ 1/4[sin3x/3] + 3[sin x] + c.
⇒ 1/4[sin3x/3] + 3sinx + c.
⇒ I = I₁ + I₂.
⇒ I = cos³x/3 - cos(x) + 1/4[sin3x/3 + 3sinx ]+ C.
⇒ I = cos³x/3 - cos(x) + sin3x/12 + 3sinx/4 + c.
MORE INFORMATION.
Basic theorem on integration.
If f(x), g(x) are two functions of a variable x and k is a constant, then
(1) = ∫K f(x)dx = K ∫f(x)dx + c.
(2) = ∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx.
(3) = d[∫f(x)dx]/dx = f(x).
(4) = ∫[d f(x)/dx]dx = f(x).