Let J=∫−5−4 3−x2tan⁡3−x2dx andK=∫−2−1 6−6x+x2tan⁡6x−x2−6dx. Then (J+K) equals __.

Solution:

We have $J = \int_{- 5}^{- 4} (3 - x^{2}) \tan (3 - x^{2}) d x$

Put $(x + 5) = t$ . Then

$\begin{array}{l} J = \int_{0}^{1} (3 - (t - 5)^{2}) \tan (3 - (t - 5)^{2}) d t \\ = \int_{0}^{1} (- 22 + 10 t - t^{2}) \tan (- 22 + 10 t - t^{2}) d t \end{array}$

Now $K = \int_{- 2}^{- 1} (6 - 6 x + x^{2}) \tan (6 x - x^{2} - 6) d x$ .

Put $(x + 2) = z$ . Then

$\begin{array}{l} K = \int_{0}^{1} (6 - 6 (z - 2) + (z - 2)^{2}) \tan (6 (z - 2) - (z - 2)^{2} - 6) d z \\ = \int_{0}^{1} (22 - 10 z + z^{2}) \tan (- 22 + 10 z - z^{2}) d z \end{array}$

Hence, $(J + K) = 0$

Let $J = \int_{- 5}^{- 4} (3 - x^{2}) \tan (3 - x^{2}) d x$ and
$K = \int_{- 2}^{- 1} (6 - 6 x + x^{2}) \tan (6 x - x^{2} - 6) d x$ . Then $(J + K)$ equals __.

Solution:

Related content