# Evaluate the following definite integral

$\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \sin x+\cos x \right)dx}$

Last updated date: 18th Mar 2023

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Hint: Try to solve this question of definite integral using the formulas of indefinite integral. Distribute the integral on the two functions and then integrate the two functions indefinitely. Then apply the limits.

Before proceeding with the question, we must know all the formulas that will be required to solve this question.

In indefinite integrals, we have a formula which we can use to integrate $\sin x$ function. That formula is,

$\int{\sin xdx=-\cos x}..............\left( 1 \right)$

Also, in indefinite integrals, we have a formula which we can use to integrate $\cos x$ function. That formula is,

$\int{\cos xdx=\sin x}..............\left( 2 \right)$

In the question, we have to evaluate the definite integral $\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \sin x+\cos x \right)dx}$. We may notice that we cannot use any property of definite integral to solve this question. Hence, we will solve this question by integrating it using the formulas of indefinite integration.

Since integral function can be distributed over addition, distributing the integral on the two functions, we get,

$\int\limits_{0}^{\dfrac{\pi }{4}}{\sin xdx}+\int\limits_{0}^{\dfrac{\pi }{4}}{\cos xdx}$

Substituting $\int{\sin xdx=-\cos x}$ from equation $\left( 1 \right)$ and $\int{\cos xdx=\sin x}$ equation $\left( 2 \right)$ in the above integration, we get,

$\left[ -\cos x \right]+\left[ \sin x \right]$

Applying limits $0$to $\dfrac{\pi }{4}$, we get

$\begin{align}

& \left[ -\cos x \right]_{0}^{{}^{\pi }/{}_{4}}+\left[ \sin x \right]_{0}^{{}^{\pi }/{}_{4}} \\

& \Rightarrow \left( -\cos \dfrac{\pi }{4}-\left( -\cos 0 \right) \right)+\left( \sin \dfrac{\pi }{4}-\sin 0 \right)...............\left( 3 \right) \\

\end{align}$

From trigonometry, we have some formulas, which are listed below,

\[\begin{align}

& \cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}.............\left( 4 \right) \\

& \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}.............\left( 5 \right) \\

& \cos 0=1...............\left( 6 \right) \\

& \sin 0=0................\left( 7 \right) \\

\end{align}\]

Substituting \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] from equation $\left( 4 \right)$, \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] from equation $\left( 5 \right)$, $\cos 0=1$ from equation $\left( 6 \right)$ and $\sin 0=0$ from equation $\left( 7 \right)$ in equation $\left( 3 \right)$, we get,

$\begin{align}

& \left( -\dfrac{1}{\sqrt{2}}-\left( -1 \right) \right)+\left( \dfrac{1}{\sqrt{2}}-0 \right) \\

& \Rightarrow \left( 1-\dfrac{1}{\sqrt{2}} \right)+\left( \dfrac{1}{\sqrt{2}} \right) \\

& \Rightarrow 1 \\

\end{align}$

Hence, the answer of the integration $\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \sin x+\cos x \right)dx}$ is $1$.

Note: There is a possibility that one may commit a mistake while writing the formula of the integral of $\sin x$. There is always a confusion regarding the sign while integrating $\sin x$. There is a possibility that one may write the formula as $\int{\sin xdx=\cos x}$ instead of $\int{\sin xdx=-\cos x}$ which may lead to an incorrect answer.

__Complete step-by-step answer:__Before proceeding with the question, we must know all the formulas that will be required to solve this question.

In indefinite integrals, we have a formula which we can use to integrate $\sin x$ function. That formula is,

$\int{\sin xdx=-\cos x}..............\left( 1 \right)$

Also, in indefinite integrals, we have a formula which we can use to integrate $\cos x$ function. That formula is,

$\int{\cos xdx=\sin x}..............\left( 2 \right)$

In the question, we have to evaluate the definite integral $\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \sin x+\cos x \right)dx}$. We may notice that we cannot use any property of definite integral to solve this question. Hence, we will solve this question by integrating it using the formulas of indefinite integration.

Since integral function can be distributed over addition, distributing the integral on the two functions, we get,

$\int\limits_{0}^{\dfrac{\pi }{4}}{\sin xdx}+\int\limits_{0}^{\dfrac{\pi }{4}}{\cos xdx}$

Substituting $\int{\sin xdx=-\cos x}$ from equation $\left( 1 \right)$ and $\int{\cos xdx=\sin x}$ equation $\left( 2 \right)$ in the above integration, we get,

$\left[ -\cos x \right]+\left[ \sin x \right]$

Applying limits $0$to $\dfrac{\pi }{4}$, we get

$\begin{align}

& \left[ -\cos x \right]_{0}^{{}^{\pi }/{}_{4}}+\left[ \sin x \right]_{0}^{{}^{\pi }/{}_{4}} \\

& \Rightarrow \left( -\cos \dfrac{\pi }{4}-\left( -\cos 0 \right) \right)+\left( \sin \dfrac{\pi }{4}-\sin 0 \right)...............\left( 3 \right) \\

\end{align}$

From trigonometry, we have some formulas, which are listed below,

\[\begin{align}

& \cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}.............\left( 4 \right) \\

& \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}.............\left( 5 \right) \\

& \cos 0=1...............\left( 6 \right) \\

& \sin 0=0................\left( 7 \right) \\

\end{align}\]

Substituting \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] from equation $\left( 4 \right)$, \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] from equation $\left( 5 \right)$, $\cos 0=1$ from equation $\left( 6 \right)$ and $\sin 0=0$ from equation $\left( 7 \right)$ in equation $\left( 3 \right)$, we get,

$\begin{align}

& \left( -\dfrac{1}{\sqrt{2}}-\left( -1 \right) \right)+\left( \dfrac{1}{\sqrt{2}}-0 \right) \\

& \Rightarrow \left( 1-\dfrac{1}{\sqrt{2}} \right)+\left( \dfrac{1}{\sqrt{2}} \right) \\

& \Rightarrow 1 \\

\end{align}$

Hence, the answer of the integration $\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \sin x+\cos x \right)dx}$ is $1$.

Note: There is a possibility that one may commit a mistake while writing the formula of the integral of $\sin x$. There is always a confusion regarding the sign while integrating $\sin x$. There is a possibility that one may write the formula as $\int{\sin xdx=\cos x}$ instead of $\int{\sin xdx=-\cos x}$ which may lead to an incorrect answer.

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