\int_{1}^{0} \tan(x^2)dx+2\int_{1}^{0} x^2 \sec^2(x^2)dx=\frac{\pi}{4} Select one: True False

rastafarral6 2021-12-02 Answered
\(\displaystyle{\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{{\sec}^{{2}}{\left({x}^{{2}}\right)}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{4}}}}\)
Select one:
True
False

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Expert Answer

Steven Arredondo
Answered 2021-12-03 Author has 7621 answers

Step1
This qestion belongs to the definite intergal in which we have to justify the given statement is true or false by
providing the correct solution of the given question. We can write the given statement below
\(\displaystyle{\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{4}}}}\) to solve the question we move to the nest step-2
Step 2
first, we write the given statement below and take the LHS side of the given statement so
\(\displaystyle{\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{4}}}}\)
now take the LHS side
\(\displaystyle{L}{H}{S}={\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
Now use the integration by part consider p and q function of x
\(\displaystyle\int{p}{q}{\left.{d}{x}\right.}={p}\int{q}{\left.{d}{x}\right.}-\int{\left({\frac{{{d}{p}}}{{{\left.{d}{x}\right.}}}}\int{q}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}\)
now apply
\(\displaystyle={\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{\int_{{{1}}}^{{{0}}}}{1}{\left.{d}{x}\right.}-{\int_{{{1}}}^{{{0}}}}{\left({\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\tan{{\left({x}^{{2}}\right)}}}{\int_{{{1}}}^{{{0}}}}{1}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{{2}}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{x}{{\mid}_{{1}}^{{0}}}-{\int_{{{1}}}^{{{0}}}}{\left({\sec{{\left({x}^{{2}}\right)}}}{\left({\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{x}^{{2}}\right)}.{x}\right)}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{x}{{\mid}_{{0}}^{{1}}}-{\int_{{{1}}}^{{{0}}}}{\left({\sec{{\left({x}^{{2}}\right)}}}{\left({2}{x}\right)}.{x}\right)}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{x}{{\mid}_{{0}}^{{1}}}-{2}{\int_{{{1}}}^{{{0}}}}{\left({\sec{{\left({x}^{{2}}\right)}}}{\left({2}{x}\right)}{x}.{x}\right)}{)}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{x}{{\mid}_{{0}}^{{1}}}-{2}{\int_{{{1}}}^{{{0}}}}{\left({\sec{{\left({x}^{{2}}\right)}}}{\left({x}^{{2}}\right)}\right)}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{x}{{\mid}_{{0}}^{{1}}}-{2}{\int_{{{1}}}^{{{0}}}}{\left({\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\right.}\)
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{x}{{\mid}_{{0}}^{{1}}}-{0}\)
now put the limits
\(\displaystyle={\tan{{\left({x}^{{2}}\right)}}}{x}{{\mid}_{{0}}^{{1}}}\)
\(\displaystyle={\tan{{\left({1}^{{2}}\right)}}}{\left({1}\right)}-{\tan{{\left({0}^{{2}}\right)}}}{\left({0}\right)}-{\tan{{\left({1}\right)}}}-{0}\)
\(\displaystyle{L}{H}{S}={\tan{{\left({1}\right)}}}\)
Now we got the LHS value and the RHS value is from the statement \(\frac{\pi}{4}\)
\(\displaystyle{\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}={\frac{{\pi}}{{{4}}}}\)
and
\(\displaystyle{L}{H}{S}={\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}={\tan{{\left({1}\right)}}}={1.5574}\)
\(\displaystyle{R}{H}{S}={\frac{{\pi}}{{{4}}}}={0.7853}\)
From the above
\(\displaystyle{L}{H}{S}\ne{R}{H}{S}\)
\(\displaystyle{\tan{{\left({1}\right)}}}\ne{\frac{{\pi}}{{{4}}}}\)
Now LHS is not equal to the RHS hence the statement is false
now we move to the next step for the result
Step 3Result
Result: from the above solution and analysis we conclude the given statement is false the LHS is not
equal to the RHS and we mentioned the result below
\(\displaystyle{\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}\ne{\frac{{\pi}}{{{4}}}}\Rightarrow\) false statement
so we calculate theture value is below
\(\displaystyle{\int_{{{1}}}^{{{0}}}}{\tan{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}+{2}{\int_{{{1}}}^{{{0}}}}{x}^{{2}}{\sec{{\left({x}^{{2}}\right)}}}{\left.{d}{x}\right.}={\tan{{\left({1}\right)}}}\Rightarrow\) true statement

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