Question

I=इंटीग्रेशन sin3 x cos 3x dx =
​

Answer 1

EXPLANATION.

⇒ ∫sin³xcos³xdx.

As we know that,

We can write equation as,

⇒ ∫sin(x).sin²xcos³xdx.

As we know that,

Formula of :

⇒ sin²x + cos²x = 1.

⇒ sin²x = 1 - cos²x.

Put the value of sin²x in equation, we get.

⇒ ∫sin(x)(1 - cos²x)cos³xdx.

By using substitution method.

Let we assume that,

⇒ cos(x) = t.

Differentiate W.R.T x, we get.

⇒ -sin(x)dx = dt.

⇒ sin(x)dx = - dt.

Substitute the value in the equation, we get.

⇒ ∫(1 - cos²x)cos³xsin(x)dx.

⇒ ∫(1 - t²)t³(-dt).

⇒ -∫(1 - t²)t³dt.

⇒ -[∫t³dt - ∫t⁵dt].

As we know that,

Formula of :

⇒ ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠-1).

Using this formula in equation, we get.

⇒ -[t³⁺¹/3 + 1 - t⁵⁺¹/5 + 1] + c.

⇒ -[t⁴/4 - t⁶/6] + c.

⇒ t⁶/6 - t⁴/4 + c.

Put the value of t = cos(x) in equation, we get.

⇒ cos⁶(x)/6 - cos⁴(x)/4 + c.

                                                                                                                             

MORE INFORMATION.

Basic theorem of 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 + c.

(3) = d(∫f(x)dx)/dx = f(x).

(4) = ∫(d(f(x)/dx)dx = f(x).

Answer 2

Correct Question:-

$\tt{Integrate \sin^{3}x \: { \cos }^{3} x \: dx}$

Solution:-

$\tt{ \int { \sin}^{3} x \: { \cos }^{3} x \: dx}$

$⟹ \tt{\int \sin \: x. { \sin }^{2} x \: \cos^{3}x \: dx}$

$⟹ \tt{\int \: \sin \: x(1 - { \cos}^{2} ) { \cos}^{3} x \: dx \: \: \: \: \: \: ( { \sin}^{2} \theta = 1 - { \cos}^{2} \theta)}$

$⟹ \tt{ \int(1 - { \cos }^{2} x) { \cos }^{3} x. \sin x \: dx} \: \: \: \: \: ....(1)$

Let cos x = t

Differentiating w.r.t.x

$\tt{ - \sin x = \frac{dt}{dx} }$

$\tt{dx = \frac{dt}{ - \sin x } }$

Thus, our equation becomes

$\tt{ \int { \sin}^{2} x \: \cos^{3} x \: dx}$

$⟹ \int{(1 - { \cos}^{2} x) { \cos}^{3} . \sin x \: dx}$

$⟹ \tt{\int(1 - {t}^{2} ) {t}^{3} . \sin x \times \frac{1}{ - \sin x} dt}$

$⟹ \tt{\int - (t - {t}^{2} ) {t}^{3} \: dt}$

$⟹\tt{ \int ({t}^{3} - {t}^{5} )dt}$

$⟹\left[ \int {t}^{3} \: dt - \int {t}^{5} \: dt\right]$

$</p><p>⟹ \tt{\left[ \frac{ {t}^{3 + 1} }{3 + 1} - \frac{ {t}^{5 + 1} }{5 + 1} \right] + c}$

$⟹ \tt{\left[{\frac{ {t}^{4} }{4}} - \frac{ {t}^{6} }{6} \right] + c}$

$⟹ \tt{\frac{ - {t }^{4} }{4} + \frac{ {t}^{6} }{6} + c }$

$⟹ \tt{\frac{ {t}^{6} }{6} - \frac{ {t}^{4} }{6} + c}$

Putting back the value of t = cos x

$⟹ \tt{\frac{ { \cos}^{6}x }{6} - \frac{ { \cos}^{4} x}{4} + c} \: \: \red{Ans.}$

I hope it's help you...☺