Question

integrate sin x + cos x by 9 + 16 sin 2x DX​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int\dfrac{\sin(x)+\cos(x)}{9+16\,\sin(2x)}\,dx$

$\displaystyle=\int\dfrac{\sin(x)+\cos(x)}{25-16+16\,\sin(2x)}\,dx$

$\displaystyle=\int\dfrac{\sin(x)+\cos(x)}{25-16\left\{1-\sin(2x)\right\}}\,dx$

$\displaystyle=\int\dfrac{\sin(x)+\cos(x)}{25-16\left\{\sin^{2}(x)+\cos^{2}(x)-2\sin(x)\cos(x)\right\}}\,dx$

$\displaystyle=\int\dfrac{\sin(x)+\cos(x)}{25-16\big\{\sin(x)-\cos(x)\big\}^{2}}\,dx$

$\bf{Put\,\,\,sin(x)-cos(x)=t}$

$\bf{\mapsto\,\,\big(cos(x)+sin(x)\big)dx=dt}$

So,

$\displaystyle=\int\dfrac{dt}{25-16\,{t}^{2}}$

$\displaystyle=\int\dfrac{dt}{\left(5\right)^{2}-\left(4t\right)^{2}}$

We know,

$\boxed{\displaystyle\bf{\int\dfrac{dx}{\left(a\right)^{2}-\left(b\,x\right)^{2}}=\dfrac{1}{b}\cdot\dfrac{1}{2a}\ln\left|\dfrac{a+b\,x}{a-b\,x}\right|+C}}$

So,

$\displaystyle=\dfrac{1}{4}\cdot\dfrac{1}{2\cdot5}\cdot\ln\left|\dfrac{5+4t}{5-4t}\right|+C$

$\displaystyle=\dfrac{1}{40}\cdot\ln\left|\dfrac{5+4t}{5-4t}\right|+C$

$\displaystyle=\dfrac{1}{40}\cdot\ln\left|\dfrac{5+4\big\{\sin(x)-\cos(x)\big\}}{5-4\big\{\sin(x)-\cos(x)\big\}}\right|+C$

$\displaystyle=\dfrac{1}{40}\cdot\ln\left|\dfrac{5+4\sin(x)-4\cos(x)}{5-4\sin(x)+4\cos(x)}\right|+C$