f(x)=1/(4sec(x)-1) −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 −3 -2.5 −2 -1.5 −1 -0.5 0 0.5 1 1.5 2 2.5 3 x y
Exercise∫ 1 4 s e c ( x ) − 1 d x \int\frac{1}{4sec\left(x\right)-1}dx ∫ 4 sec ( x ) − 1 1 d x
Step-by-step Solution
1
We can solve the integral ∫ 1 4 sec ( x ) − 1 d x \int\frac{1}{4\sec\left(x\right)-1}dx ∫ 4 s e c ( x ) − 1 1 d x t t t
t = tan ( x 2 ) t=\tan\left(\frac{x}{2}\right) t = tan ( 2 x )
sin x = 2 t 1 + t 2 , cos x = 1 − t 2 1 + t 2 , a n d d x = 2 1 + t 2 d t \sin x=\frac{2t}{1+t^{2}},\:\cos x=\frac{1-t^{2}}{1+t^{2}},\:\mathrm{and}\:\:dx=\frac{2}{1+t^{2}}dt sin x = 1 + t 2 2 t , cos x = 1 + t 2 1 − t 2 , and d x = 1 + t 2 2 d t
3
Substituting in the original integral we get
∫ 1 4 ( 1 + t 2 1 − t 2 ) − 1 2 1 + t 2 d t \int\frac{1}{4\left(\frac{1+t^{2}}{1-t^{2}}\right)-1}\frac{2}{1+t^{2}}dt ∫ 4 ( 1 − t 2 1 + t 2 ) − 1 1 1 + t 2 2 d t
Intermediate steps
∫ 2 ( 1 − t 2 ) ( 4 ( 1 + t 2 ) − ( 1 − t 2 ) ) ( 1 + t 2 ) d t \int\frac{2\left(1-t^{2}\right)}{\left(4\left(1+t^{2}\right)-\left(1-t^{2}\right)\right)\left(1+t^{2}\right)}dt ∫ ( 4 ( 1 + t 2 ) − ( 1 − t 2 ) ) ( 1 + t 2 ) 2 ( 1 − t 2 ) d t
Explain this step further
5
Take out the constant 2 2 2
2 ∫ 1 − t 2 ( 4 ( 1 + t 2 ) − ( 1 − t 2 ) ) ( 1 + t 2 ) d t 2\int\frac{1-t^{2}}{\left(4\left(1+t^{2}\right)-\left(1-t^{2}\right)\right)\left(1+t^{2}\right)}dt 2 ∫ ( 4 ( 1 + t 2 ) − ( 1 − t 2 ) ) ( 1 + t 2 ) 1 − t 2 d t
6
Solve the product 4 ( 1 + t 2 ) 4\left(1+t^{2}\right) 4 ( 1 + t 2 )
2 ∫ 1 − t 2 ( 4 + 4 t 2 − ( 1 − t 2 ) ) ( 1 + t 2 ) d t 2\int\frac{1-t^{2}}{\left(4+4t^{2}-\left(1-t^{2}\right)\right)\left(1+t^{2}\right)}dt 2 ∫ ( 4 + 4 t 2 − ( 1 − t 2 ) ) ( 1 + t 2 ) 1 − t 2 d t
7
Solve the product − ( 1 − t 2 ) -\left(1-t^{2}\right) − ( 1 − t 2 )
2 ∫ 1 − t 2 ( 4 + 4 t 2 − 1 + t 2 ) ( 1 + t 2 ) d t 2\int\frac{1-t^{2}}{\left(4+4t^{2}-1+t^{2}\right)\left(1+t^{2}\right)}dt 2 ∫ ( 4 + 4 t 2 − 1 + t 2 ) ( 1 + t 2 ) 1 − t 2 d t
Intermediate steps
8
Simplify the expression
2 ∫ 1 − t 2 ( 3 + 5 t 2 ) ( 1 + t 2 ) d t 2\int\frac{1-t^{2}}{\left(3+5t^{2}\right)\left(1+t^{2}\right)}dt 2 ∫ ( 3 + 5 t 2 ) ( 1 + t 2 ) 1 − t 2 d t
Explain this step further
Intermediate steps
9
Rewrite the fraction 1 − t 2 ( 3 + 5 t 2 ) ( 1 + t 2 ) \frac{1-t^{2}}{\left(3+5t^{2}\right)\left(1+t^{2}\right)} ( 3 + 5 t 2 ) ( 1 + t 2 ) 1 − t 2 2 2 2
4 3 + 5 t 2 + − 1 1 + t 2 \frac{4}{3+5t^{2}}+\frac{-1}{1+t^{2}} 3 + 5 t 2 4 + 1 + t 2 − 1
Explain this step further
10
Expand the integral ∫ ( 4 3 + 5 t 2 + − 1 1 + t 2 ) d t \int\left(\frac{4}{3+5t^{2}}+\frac{-1}{1+t^{2}}\right)dt ∫ ( 3 + 5 t 2 4 + 1 + t 2 − 1 ) d t 2 2 2
2 ∫ 4 3 + 5 t 2 d t + 2 ∫ − 1 1 + t 2 d t 2\int\frac{4}{3+5t^{2}}dt+2\int\frac{-1}{1+t^{2}}dt 2 ∫ 3 + 5 t 2 4 d t + 2 ∫ 1 + t 2 − 1 d t
Intermediate steps
11
The integral 2 ∫ 4 3 + 5 t 2 d t 2\int\frac{4}{3+5t^{2}}dt 2 ∫ 3 + 5 t 2 4 d t 8 3 5 arctan ( 5 3 t ) 3 \frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}t\right)}{3} 3 8 5 3 a r c t a n ( 3 5 t )
8 3 5 arctan ( 5 3 t ) 3 \frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}t\right)}{3} 3 8 5 3 arctan ( 3 5 t )
Explain this step further
Intermediate steps
12
The integral 2 ∫ − 1 1 + t 2 d t 2\int\frac{-1}{1+t^{2}}dt 2 ∫ 1 + t 2 − 1 d t − 2 arctan ( t ) -2\arctan\left(t\right) − 2 arctan ( t )
− 2 arctan ( t ) -2\arctan\left(t\right) − 2 arctan ( t )
Explain this step further
13
Gather the results of all integrals
8 3 5 arctan ( 5 3 t ) 3 − 2 arctan ( t ) \frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}t\right)}{3}-2\arctan\left(t\right) 3 8 5 3 arctan ( 3 5 t ) − 2 arctan ( t )
14
Replace t t t tan ( x 2 ) \tan\left(\frac{x}{2}\right) tan ( 2 x )
8 3 5 arctan ( 5 3 tan ( x 2 ) ) 3 − 2 arctan ( tan ( x 2 ) ) \frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-2\arctan\left(\tan\left(\frac{x}{2}\right)\right) 3 8 5 3 arctan ( 3 5 tan ( 2 x ) ) − 2 arctan ( tan ( 2 x ) )
Intermediate steps
15
Simplify the expression
8 3 5 arctan ( 5 3 tan ( x 2 ) ) 3 − x \frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-x 3 8 5 3 arctan ( 3 5 tan ( 2 x ) ) − x
Explain this step further
16
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration C C C
8 3 5 arctan ( 5 3 tan ( x 2 ) ) 3 − x + C 0 \frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-x+C_0 3 8 5 3 arctan ( 3 5 tan ( 2 x ) ) − x + C 0
Final answer to the exercise 8 3 5 arctan ( 5 3 tan ( x 2 ) ) 3 − x + C 0 \frac{8\sqrt{\frac{3}{5}}\arctan\left(\sqrt{\frac{5}{3}}\tan\left(\frac{x}{2}\right)\right)}{3}-x+C_0 3 8 5 3 arctan ( 3 5 tan ( 2 x ) ) − x + C 0