$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$ is equal to:
A. $\pi $
B. $\dfrac{\pi }{2}$
C. $\dfrac{\pi }{3}$
D. $\dfrac{\pi }{4}$
Answer
362.7k+ views
Hint: For$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$ , use $\tan \theta
=\dfrac{\sin \theta }{\cos \theta }$, then multiply and divide by $2$and simplify. After that, split the term and use $\sin \theta +\cos \theta =u$and apply the limits. Simplify it, you will get the answer.
Complete step by step solution:
We have to integrate,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$
We know the identity, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$,
So substituting above we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}=\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{d\theta }{1+\dfrac{\sin \theta }{\cos \theta }}}$
Also, multiplying and dividing by $2$ we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}=\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}$
So we can write,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{(\cos \theta +\sin \theta +\cos \theta -\sin \theta
)d\theta }{\cos \theta +\sin \theta }}\]
Now, splitting we get,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{d\theta +\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{(\cos \theta -\sin \theta )d\theta }{\cos \theta +\sin \theta }}}\]
So let, $\sin \theta +\cos \theta =u$
Now differentiating both sides we get,
$(\cos \theta -\sin \theta )d\theta =du$
For $\theta =\dfrac{\pi }{2}$, $u=1$ and $\theta =0$, $u=1$
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{d\theta +\dfrac{1}{2}\int\limits_{1}^{1}{\dfrac{du}{u}}}\]
Now we know that, $\int{\dfrac{1}{u}=\log u+c}$
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\left[ \theta \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ \log u \right]_{1}^{1}\]
Now, applying the limit we get,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\left( \dfrac{\pi }{2}-0 \right)+\dfrac{1}{2}\left( \log 1-\log 1 \right)=\dfrac{\pi
}{4}+0=\dfrac{\pi }{4}\]
We get,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{\pi }{4}\]
We get the answer as option(D).
Note: Read the question carefully. You must be familiar with the concept of integration. Also, don’t make silly mistakes. While simplifying, take care that no term is missing. Also, take care of signs. Most of the mistakes occur while simplifying so avoid it.
=\dfrac{\sin \theta }{\cos \theta }$, then multiply and divide by $2$and simplify. After that, split the term and use $\sin \theta +\cos \theta =u$and apply the limits. Simplify it, you will get the answer.
Complete step by step solution:
We have to integrate,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$
We know the identity, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$,
So substituting above we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}=\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{d\theta }{1+\dfrac{\sin \theta }{\cos \theta }}}$
Also, multiplying and dividing by $2$ we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}=\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}$
So we can write,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{(\cos \theta +\sin \theta +\cos \theta -\sin \theta
)d\theta }{\cos \theta +\sin \theta }}\]
Now, splitting we get,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{d\theta +\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi
}{2}}{\dfrac{(\cos \theta -\sin \theta )d\theta }{\cos \theta +\sin \theta }}}\]
So let, $\sin \theta +\cos \theta =u$
Now differentiating both sides we get,
$(\cos \theta -\sin \theta )d\theta =du$
For $\theta =\dfrac{\pi }{2}$, $u=1$ and $\theta =0$, $u=1$
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{d\theta +\dfrac{1}{2}\int\limits_{1}^{1}{\dfrac{du}{u}}}\]
Now we know that, $\int{\dfrac{1}{u}=\log u+c}$
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\left[ \theta \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ \log u \right]_{1}^{1}\]
Now, applying the limit we get,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{1}{2}\left( \dfrac{\pi }{2}-0 \right)+\dfrac{1}{2}\left( \log 1-\log 1 \right)=\dfrac{\pi
}{4}+0=\dfrac{\pi }{4}\]
We get,
\[\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta
}}=\dfrac{\pi }{4}\]
We get the answer as option(D).
Note: Read the question carefully. You must be familiar with the concept of integration. Also, don’t make silly mistakes. While simplifying, take care that no term is missing. Also, take care of signs. Most of the mistakes occur while simplifying so avoid it.
Last updated date: 01st Oct 2023
•
Total views: 362.7k
•
Views today: 9.62k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

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

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Why are resources distributed unequally over the e class 7 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

What is the past tense of read class 10 english CBSE
