Trigonometric Integral

Calculus Level 4

∫ cos ⁡ x − sin ⁡ x ( 1 + cos ⁡ x ) cos ⁡ x + cos ⁡ x sin ⁡ x + ( 1 + sin ⁡ x ) sin ⁡ x d x = . . . \int\frac{\cos x-\sin x}{(1+\cos x)\cos x+\cos x\sin x+(1+\sin x)\sin x}dx=\;...

1 − tan ⁡ 1 2 x 1 + cot ⁡ 1 2 x + C \frac{1-\tan\frac{1}{2}x}{1+\cot\frac{1}{2}x}+\text{C} sin ⁡ 1 2 x + cos ⁡ 1 2 x sin ⁡ 1 2 x − cos ⁡ 1 2 x + C \frac{\sin\frac{1}{2}x+\cos\frac{1}{2}x}{\sin\frac{1}{2}x-\cos\frac{1}{2}x}+\text{C} cos ⁡ 1 2 x − sin ⁡ 1 2 x sin ⁡ 1 2 x + cos ⁡ 1 2 x + C \frac{\cos\frac{1}{2}x-\sin\frac{1}{2}x}{\sin\frac{1}{2}x+\cos\frac{1}{2}x}+\text{C} 1 + cot ⁡ 1 2 x 1 + tan ⁡ 1 2 x + C \frac{1+\cot\frac{1}{2}x}{1+\tan\frac{1}{2}x}+\text{C}

Tunk-Fey Ariawan by Tunk-Fey Ariawan , 35, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tunk-Fey Ariawan
Mar 17, 2014

First, we simplify the trigonometric function. cos ⁡ x − sin ⁡ x ( 1 + cos ⁡ x ) cos ⁡ x + cos ⁡ x sin ⁡ x + ( 1 + sin ⁡ x ) sin ⁡ x = 1 + cos ⁡ x − 1 − sin ⁡ x ) cos ⁡ x + cos ⁡ 2 x + cos ⁡ x sin ⁡ x + sin ⁡ x + sin ⁡ 2 x = 1 + cos ⁡ x − ( 1 + sin ⁡ x ) sin ⁡ 2 x + cos ⁡ 2 x + sin ⁡ x + sin ⁡ x cos ⁡ x + cos ⁡ x = 1 + cos ⁡ x − ( 1 + sin ⁡ x ) 1 + sin ⁡ x + sin ⁡ x cos ⁡ x + cos ⁡ x = 1 1 + sin ⁡ x − 1 1 + cos ⁡ x = 1 1 + sin ⁡ x ⋅ 1 − sin ⁡ x 1 − sin ⁡ x − 1 sin ⁡ 2 1 2 x + cos ⁡ 2 1 2 x + cos ⁡ 2 1 2 x − sin ⁡ 2 1 2 x = 1 − sin ⁡ x 1 − sin ⁡ 2 x − 1 2 cos ⁡ 2 1 2 x = 1 − sin ⁡ x cos ⁡ 2 x − 1 2 cos ⁡ 2 1 2 x = 1 cos ⁡ 2 x − sin ⁡ x cos ⁡ 2 x − 1 2 cos ⁡ 2 1 2 x . \begin{aligned} \frac{\cos x-\sin x}{(1+\cos x)\cos x+\cos x\sin x+(1+\sin x)\sin x}&=\frac{1+\cos x-1-\sin x)}{\cos x+\cos^2 x+\cos x\sin x+\sin x+\sin^2 x}\\ &=\frac{1+\cos x-(1+\sin x)}{\sin^2 x+\cos^2 x+\sin x+\sin x\cos x+\cos x}\\ &=\frac{1+\cos x-(1+\sin x)}{1+\sin x+\sin x\cos x+\cos x}\\ &=\frac{1}{1+\sin x}-\frac{1}{1+\cos x}\\ &=\frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}-\frac{1}{\sin^2\frac{1}{2}x+\cos^2\frac{1}{2}x+\cos^2\frac{1}{2}x-\sin^2\frac{1}{2}x}\\ &=\frac{1-\sin x}{1-\sin^2 x}-\frac{1}{2\cos^2\frac{1}{2}x}\\ &=\frac{1-\sin x}{\cos^2 x}-\frac{1}{2\cos^2\frac{1}{2}x}\\ &=\frac{1}{\cos^2 x}-\frac{\sin x}{\cos^2 x}-\frac{1}{2\cos^2\frac{1}{2}x}.\\ \end{aligned} Thus ∫ cos ⁡ x − sin ⁡ x ( 1 + cos ⁡ x ) cos ⁡ x + cos ⁡ x sin ⁡ x + ( 1 + sin ⁡ x ) sin ⁡ x d x = ∫ ( 1 cos ⁡ 2 x − sin ⁡ x cos ⁡ 2 x − 1 2 cos ⁡ 2 1 2 x ) d x = ∫ 1 cos ⁡ 2 x d x − ∫ sin ⁡ x cos ⁡ 2 x d x − ∫ 1 2 cos ⁡ 2 1 2 x d x = ∫ d ( tan ⁡ x ) + ∫ 1 cos ⁡ 2 x d ( cos ⁡ x ) − ∫ d ( tan ⁡ 1 2 x ) = tan ⁡ x − 1 cos ⁡ x − tan ⁡ 1 2 x + C = sin ⁡ x cos ⁡ x − 1 cos ⁡ x − tan ⁡ 1 2 x + C = − 1 − sin ⁡ x cos ⁡ x − tan ⁡ 1 2 x + C = − sin ⁡ 2 1 2 x + cos ⁡ 2 1 2 x − 2 sin ⁡ 1 2 x cos ⁡ 1 2 x cos ⁡ 2 1 2 x − sin ⁡ 2 1 2 x − tan ⁡ 1 2 x + C = ( sin ⁡ 1 2 x − cos ⁡ 1 2 x ) 2 ( sin ⁡ 1 2 x − cos ⁡ 1 2 x ) ( sin ⁡ 1 2 x + cos ⁡ 1 2 x ) − tan ⁡ 1 2 x + C = sin ⁡ 1 2 x − cos ⁡ 1 2 x sin ⁡ 1 2 x + cos ⁡ 1 2 x − tan ⁡ 1 2 x + C = sin ⁡ 1 2 x ( 1 − cot ⁡ 1 2 x ) sin ⁡ 1 2 x ( 1 + cot ⁡ 1 2 x ) − tan ⁡ 1 2 x + ( C + 1 ) = 1 − cot ⁡ 1 2 x 1 + cot ⁡ 1 2 x + ( 1 − tan ⁡ 1 2 x ) + C = 1 − cot ⁡ 1 2 x + ( 1 − tan ⁡ 1 2 x ) ( 1 + cot ⁡ 1 2 x ) 1 + cot ⁡ 1 2 x + C = 1 − cot ⁡ 1 2 x + 1 + cot ⁡ 1 2 x − tan ⁡ 1 2 x − 1 1 + cot ⁡ 1 2 x + C = 1 − tan ⁡ 1 2 x 1 + cot ⁡ 1 2 x + C \begin{aligned} \int\frac{\cos x-\sin x}{(1+\cos x)\cos x+\cos x\sin x+(1+\sin x)\sin x}dx&=\int\left(\frac{1}{\cos^2 x}-\frac{\sin x}{\cos^2 x}-\frac{1}{2\cos^2\frac{1}{2}x}\right)dx\\ &=\int\frac{1}{\cos^2 x}dx-\int\frac{\sin x}{\cos^2 x}dx-\int\frac{1}{2\cos^2\frac{1}{2}x}dx\\ &=\int\;d(\tan x)+\int\frac{1}{\cos^2 x}d(\cos x)-\int\;d\left(\tan\frac{1}{2}x\right)\\ &=\tan x-\frac{1}{\cos x}-\tan\frac{1}{2}x+\text{C}\\ &=\frac{\sin x}{\cos x}-\frac{1}{\cos x}-\tan\frac{1}{2}x+\text{C}\\ &=-\frac{1-\sin x}{\cos x}-\tan\frac{1}{2}x+\text{C}\\ &=-\frac{\sin^2\frac{1}{2}x+\cos^2\frac{1}{2}x-2\sin\frac{1}{2}x\cos\frac{1}{2}x}{\cos^2\frac{1}{2}x-\sin^2\frac{1}{2}x}-\tan\frac{1}{2}x+\text{C}\\ &=\frac{\left(\sin\frac{1}{2}x-\cos\frac{1}{2}x\right)^2}{\left(\sin\frac{1}{2}x-\cos\frac{1}{2}x\right)\left(\sin\frac{1}{2}x+\cos\frac{1}{2}x\right)}-\tan\frac{1}{2}x+\text{C}\\ &=\frac{\sin\frac{1}{2}x-\cos\frac{1}{2}x}{\sin\frac{1}{2}x+\cos\frac{1}{2}x}-\tan\frac{1}{2}x+\text{C}\\ &=\frac{\sin\frac{1}{2}x\left(1-\cot\frac{1}{2}x\right)}{\sin\frac{1}{2}x\left(1+\cot\frac{1}{2}x\right)}-\tan\frac{1}{2}x+(\text{C}+1)\\ &=\frac{1-\cot\frac{1}{2}x}{1+\cot\frac{1}{2}x}+\left(1-\tan\frac{1}{2}x\right)+\text{C}\\ &=\frac{1-\cot\frac{1}{2}x+\left(1-\tan\frac{1}{2}x\right)\left(1+\cot\frac{1}{2}x\right)}{1+\cot\frac{1}{2}x}+\text{C}\\ &=\frac{1-\cot\frac{1}{2}x+1+\cot\frac{1}{2}x-\tan\frac{1}{2}x-1}{1+\cot\frac{1}{2}x}+\text{C}\\ &=\frac{1-\tan\frac{1}{2}x}{1+\cot\frac{1}{2}x}+\text{C}\\ \end{aligned} Another way to solve ∫ cos ⁡ x − sin ⁡ x ( 1 + cos ⁡ x ) cos ⁡ x + cos ⁡ x sin ⁡ x + ( 1 + sin ⁡ x ) sin ⁡ x d x = ∫ ( 1 1 + sin ⁡ x − 1 1 + cos ⁡ x ) d x \int\frac{\cos x-\sin x}{(1+\cos x)\cos x+\cos x\sin x+(1+\sin x)\sin x}dx=\int\left(\frac{1}{1+\sin x}-\frac{1}{1+\cos x}\right)dx is using Weierstrass substitution . By using Weierstrass substitution, the integral can be quickly obtained.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...