Question

integration of √1+sin5x. dx​

Answer 1

EXPLANATION.

⇒ ∫√1 + sin5x dx.

As we know that,

Formula of :

sin²x + cos²x = 1.

sin2x = 2sinxcosx.

Using this formula in the equation, we get.

We can write sin5x as,

⇒ sin5x = 2sin(5x/2).cos(5x/2).

Now, we can write equation as,

⇒ ∫[√sin²(5x/2) + cos²(5x/2) + 2sin(5x/2).cos(5x/2)]dx.

⇒ ∫√[sin(5x/2) + cos(5x/2)]²dx.

⇒ ∫[sin(5x/2) + cos(5x/2)]dx.

⇒ ∫sin(5x/2)dx + ∫cos(5x/2)dx.

⇒ - cos(5x/2)/(5/2) + sin(5x/2)/(5/2) + c.

⇒ -2/5 cos(5x/2) + 2/5 sin(5x/2) + c.

⇒ 2/5[sin(5x/2) - cos(5x/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) ∫dx/x = ㏒(x) + c.

(6) ∫eˣdx = eˣ + c.

(7) ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.

Answer 2

The Final Answer is

$\boxed{ \color{lime} \rm \frac{2}{5} [ \: \sin( \frac{5x}{2}) - \cos( \frac{5x}{2} ) \: ] \: + \: c}$

For Further Explanation,

Refer the given attachment