Answer

Verified

451.5k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Guru Purnima speech in English in 100 words class 7 english CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Select the word that is correctly spelled a Twelveth class 10 english CBSE