Question

integrate : 1/(2sinx+3cosx)^2"dx

Answer 1

Answer:

-1/(4tanx +6)

Step-by-step explanation:

∫1/(2sinx+3cosx)^2 dx

Dividing both Numerator and denominator by cos²x, we get

∫ (1/cos²x)/(2sinx+3cosx)^2/cos²x dx

∫ sec²x/(2tan x + 3)² dx----(*)

Let z = 2tanx + 3

Differentiating on both sides we get,

dz = 2 sec²x dx

=> sec²x dx = dz/2--(1)

Replacing (1) in (*), we get

∫ 1/(2z²) dz

∫ z⁻² /2 dz

We know that ∫ xⁿ dx = xⁿ⁺¹/n+1

=> ∫ z⁻² /2 dz

= 1/2*z⁻²⁺¹/-2+1 +c

=1/2*z⁻¹/-1 +c

-1/2z +c

Substituting back z = 2tanx + 3, we get

= -1/2(2tanx + 3) + c where c is an arbitrary constant

Answer 2

Answer:
$\int \frac{1}{ {(2 \: sin \: x + 3 \: cos \: x)}^{2} } dx= \frac{ - 1}{2(2 \: tan \: x + 3)} + c$

Solution:

$=>\int \frac{1}{ {(2 \: sin \: x + 3 \: cos \: x)}^{2} } dx \\ \\ => \int\frac{1}{ {cos}^{2} x( {2 \: tan \: x + 3)}^{2} } dx \\ \\ => \int \frac{ {sec}^{2} x}{( {2tan \: x + 3)}^{2} } dx \\ \\ let \: \: tan \: x = t \\ \\ {sec}^{2} x dx= dt \\ \\ => \int\frac{1}{ {(2t + 3)}^{2} } dt$

let

$2t = u \\ \\ 2 \: dt = du \\ \\ dt = \frac{du}{2} \\ \\ => \int\frac{1}{2( {u + 3)}^{2} }du \\ \\ - \frac{1}{2(u + 3)} + c$

redo substitution

$\int \frac{1}{ {(2 \: sin \: x + 3 \: cos \: x)}^{2} } dx= \frac{ - 1}{2(2 \: tan \: x + 3)} + c$