Question

How do I solve for k in an equation with integrals?

Answer 1

In this question, we have to find how to solve the constant with the integrals. Thus, we will apply the integral formula to get the result. First, we will let a definite integral, where the limits of the function is from 0 to k. Then, we will apply the integral formula in the function by making the necessary mathematical calculations. In the end, we will apply the integral formula $\int\limits_{a}^{b}{f(x)dx=f(b)=f(a)}$ in the equation. Then, we will substitute the value of integral equal to the answer of the integral, to get the required result for the problem.

Complete step by step solution:
According to the problem, we have to find the value of k for a definite integral. As we know, definite integral involves the limit of the function.
Let us suppose a definite integral $\int\limits_{0}^{k}{\dfrac{{{\sec }^{2}}x}{1+\tan x}dx}$ which is equal to log(2) -------- (1)
Now, we will apply the integral formula in the above integral. First, we will apply the substitution method by letting $1+\tan x=u$ -------- (2)
Now, we will differentiate equation (2), we get
$\Rightarrow {{\sec }^{2}}xdx=du$
Now, we will divide ${{\sec }^{2}}x$ on both sides in the above equation, we get
$\Rightarrow \dfrac{{{\sec }^{2}}x}{{{\sec }^{2}}x}dx=\dfrac{du}{{{\sec }^{2}}x}$
On further solving, we get
$\Rightarrow dx=\dfrac{du}{{{\sec }^{2}}x}$ --------- (3)
Now, we will substitute the value of equation (3) in equation (1), we get
$\Rightarrow \int\limits_{{}}^{{}}{\dfrac{{{\sec }^{2}}x}{u}\dfrac{du}{{{\sec }^{2}}x}}=\log 2$
Thus, we will not put the value of limits in the integral, because we have changed the variable. Thus, on further simplifying the above equation, we get
$\Rightarrow \int\limits_{{}}^{{}}{\dfrac{1}{u}du}=\log 2$
As we know, the integral of $\dfrac{1}{x}$ is log x, therefore we get
$\Rightarrow \log u=\log 2$ ---------- (4)
Now, we will substitute the value of equation (2) in the above equation, we get
$\Rightarrow \log \left( 1+\tan x \right)=\log 2$
Thus, we get the same variable as before, so now we will apply the limits in the above equation, we get
$\Rightarrow \left[ \log \left( 1+\tan x \right) \right]_{0}^{k}=\log 2$
Now, we will apply the integral formula $\int\limits_{a}^{b}{f(x)dx=f(b)=f(a)}$ in the above equation, we get
$\Rightarrow \log \left( 1+\tan k \right)-\log \left( 1+\tan 0 \right)=\log 2$
Now, we know that $\tan 0=0$ , thus we get
$\Rightarrow \log \left( 1+\tan k \right)-\log \left( 1 \right)=\log 2$
Also, the value of $\log 1=0$ , therefore we get
$\Rightarrow \log \left( 1+\tan k \right)=\log 2$
As we see, both the sides of the above equation has log function, thus we will remove the log function, we get
$\Rightarrow 1+\tan k=2$
Now, we will subtract 1on both sides in the above equation, we get
$\Rightarrow 1+\tan k-1=2-1$
As we know, the same terms with opposite signs cancel out each other, thus we get
$\Rightarrow \tan k=1$
Also, we know that the value of tangent function is 1 when the angle is $\dfrac{\pi }{4}+n\pi$ , thus we get
$\Rightarrow k=\dfrac{\pi }{4}+n\pi$ which is the required answer.

Note:
While solving this problem, do mention all the steps properly to avoid mathematical error. Always remember that k can be the upper limit or the lower limit of the integral, and we have to apply the formula $\int\limits_{a}^{b}{f(x)dx=f(b)=f(a)}$ , to get the solution.